Samacheer Kalvi Class 11 Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Exercise 4.2

Question 1. If \( (n - 1)P_3 : nP_4 = 1 : 10 \) find n.
Answer: Given the ratio \( (n - 1)P_3 : nP_4 = 1 : 10 \). We use the permutation formula \( nP_r = \frac{n!}{(n-r)!} \).
So, we can write the given ratio as:
\( \frac{(n - 1)!}{(n - 1 - 3)!} : \frac{n!}{(n - 4)!} = 1 : 10 \)
\( \frac{(n - 1)!}{(n - 4)!} \times \frac{(n - 4)!}{n!} = \frac{1}{10} \)
The \( (n - 4)! \) terms cancel out, leaving:
\( \frac{(n - 1)!}{n!} = \frac{1}{10} \)
Since \( n! = n \times (n - 1)! \), we can simplify the left side:
\( \frac{(n - 1)!}{n \times (n - 1)!} = \frac{1}{10} \)
\( \frac{1}{n} = \frac{1}{10} \)
This means that \( n \) must be equal to 10. Understanding permutation ratios can simplify complex problems by canceling common factorial terms.
In simple words: We are given a relationship between two permutations. By using the formula for permutations and simplifying the equation, we can find the value of 'n'.

🎯 Exam Tip: Remember the basic definition of permutations and factorial properties. \( n! = n \times (n-1)! \) is a key identity for simplifying such expressions.

Question 2. If \( 10P_{r-1} = 2 \times 6P_r \), find r.
Answer: We are given the equation \( 10P_{r-1} = 2 \times 6P_r \). We will use the permutation formula \( nP_r = \frac{n!}{(n-r)!} \).
Substitute the values into the formula:
\( \frac{10!}{(10 - (r-1))!} = 2 \times \frac{6!}{(6 - r)!} \)
\( \frac{10!}{(10 - r + 1)!} = 2 \times \frac{6!}{(6 - r)!} \)
\( \frac{10!}{(11 - r)!} = 2 \times \frac{6!}{(6 - r)!} \)
Now, we expand \( 10! \) and \( (11 - r)! \) until we find common terms to cancel:
\( \frac{10 \times 9 \times 8 \times 7 \times 6!}{(11 - r)(10 - r)(9 - r)(8 - r)(7 - r)(6 - r)!} = 2 \times \frac{6!}{(6 - r)!} \)
Cancel \( 6! \) and \( (6 - r)! \) from both sides:
\( \frac{10 \times 9 \times 8 \times 7}{(11 - r)(10 - r)(9 - r)(8 - r)(7 - r)} = 2 \)
Simplify the left side:
\( \frac{5040}{(11 - r)(10 - r)(9 - r)(8 - r)(7 - r)} = 2 \)
\( \frac{2520}{(11 - r)(10 - r)(9 - r)(8 - r)(7 - r)} = 1 \)
We need to find an integer value of \( r \) that satisfies this equation. We can check integer values, starting from small numbers, or compare factors. The product of five consecutive integers on the denominator must be 2520.
Let's try to find five consecutive integers whose product is 2520.
We see that \( 7 \times 6 \times 5 \times 4 \times 3 = 2520 \).
So, we can equate the denominators:
\( (11 - r)(10 - r)(9 - r)(8 - r)(7 - r) = 7 \times 6 \times 5 \times 4 \times 3 \)
Comparing the terms, we can see that if \( 7 - r = 3 \), then \( r = 4 \).
Let's verify: if \( r = 4 \), then the terms are \( (11 - 4)(10 - 4)(9 - 4)(8 - 4)(7 - 4) = 7 \times 6 \times 5 \times 4 \times 3 \).
Since this matches, the value of \( r \) is 4. Permutation problems often involve algebraic manipulation of factorials to find unknown variables. It's a common technique in combinatorics.
In simple words: We used the permutation formula to set up an equation. After simplifying the factorials on both sides, we found that the product of five consecutive numbers was equal to 2520. By matching these numbers, we figured out that 'r' is 4.

🎯 Exam Tip: When solving permutation equations, expanding factorials and simplifying terms is key. Often, the resulting equation can be solved by comparing products of consecutive integers.

Question 3.
(i) Suppose 8 people enter an event in a swimming meet. In how many ways could the gold, silver and bronze prizes be awarded?
(ii) Three men have 4 mats, 5 waist coats and 6 caps.In how many ways can they wear them?
Answer:
(i) For the gold, silver, and bronze prizes, we need to choose 3 people from 8 and arrange them in order because the prizes are distinct (gold is different from silver, etc.). This is a permutation problem.
The number of ways to award these prizes is given by \( 8P_3 \).
\( 8P_3 = 8 \times 7 \times 6 \)
\( 8P_3 = 336 \)
So, there are 336 ways to award the gold, silver, and bronze prizes. The order in which the prizes are given matters, hence we use permutations.
(ii) We need to find the number of ways three men can wear 4 mats, 5 waistcoats, and 6 caps. This means we are selecting and arranging items for the men.
Number of men = 3
Number of mats = 4
Number of waistcoats = 5
Number of caps = 6

For the mats, 4 mats can be given to 3 men in \( 4P_3 \) ways.
\( 4P_3 = 4 \times 3 \times 2 = 24 \) ways

For the waistcoats, 5 waistcoats can be given to 3 men in \( 5P_3 \) ways.
\( 5P_3 = 5 \times 4 \times 3 = 60 \) ways

For the caps, 6 caps can be given to 3 men in \( 6P_3 \) ways.
\( 6P_3 = 6 \times 5 \times 4 = 120 \) ways

To find the total number of ways the men can wear all three items, we multiply the ways for each item (by the fundamental principle of counting):
Total number of ways = \( 4P_3 \times 5P_3 \times 6P_3 \)
Total number of ways = \( 24 \times 60 \times 120 \)
Total number of ways = \( 1,72,800 \)
So, there are 172,800 ways for three men to wear the given items. This type of problem combines multiple permutation scenarios using the multiplication rule.
In simple words: (i) To give out gold, silver, and bronze prizes, we find the number of ways to pick 3 people from 8 and arrange them. (ii) For the clothes, we figure out how many ways to give mats to the men, then waistcoats, then caps, and multiply all those numbers together to get the total ways they can dress up.

🎯 Exam Tip: When solving permutation problems, always consider if the order of selection matters (permutations) or not (combinations). For independent events, multiply the number of ways for each event to find the total.

Question 4. Determine the number of permutations of the letters of the word SIMPLE If all are taken at a time?
Answer: The given word is SIMPLE.
First, count the total number of letters in the word SIMPLE. There are 6 letters (S, I, M, P, L, E).
Next, check if any letters are repeated. In the word SIMPLE, all the letters are distinct, meaning none of them are repeated.
When all letters are distinct and we want to arrange all of them at a time, the number of permutations is given by \( n! \), where \( n \) is the number of letters.
So, the number of ways to arrange the letters of SIMPLE is \( 6! \).
\( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
\( 6! = 720 \)
Therefore, there are 720 distinct permutations (or words) that can be formed from the letters of the word SIMPLE when all letters are taken at a time. The fundamental counting principle applies when there are no repeated items.
In simple words: The word SIMPLE has 6 different letters. To find how many ways we can arrange all these letters, we calculate 6 factorial (6!). This means multiplying all whole numbers from 6 down to 1.

🎯 Exam Tip: For permutations where all letters are distinct and all are used, the number of arrangements is simply the factorial of the number of letters. Be careful to check for repeated letters if they exist.

Question 5. A test consists of 10 multiple choice questions. In how many ways can the test be answered if
(i) Each question has four choices ?
(ii) The first four questions have three choices and the remaining have five choices?
(iii) Question number n has n + 1 choices ?
Answer:
(i) If each question has four choices, and there are 10 questions, then for each question, there are 4 independent ways to answer it. We use the multiplication principle.
Number of questions = 10
Choices per question = 4
Total ways to answer = \( 4 \times 4 \times \dots \) (10 times) \( = 4^{10} \)
So, there are \( 4^{10} \) ways to answer the test. This is a common application of the exponent rule in counting problems.

(ii) Here, the number of choices varies for different questions. We apply the multiplication principle based on these changes.
For the first four questions: Each has 3 choices.
Ways for first 4 questions = \( 3 \times 3 \times 3 \times 3 = 3^4 \)
For the remaining 6 questions (from 5 to 10): Each has 5 choices.
Ways for remaining 6 questions = \( 5 \times 5 \times 5 \times 5 \times 5 \times 5 = 5^6 \)
Total ways to answer the test = (Ways for first 4 questions) \( \times \) (Ways for remaining 6 questions)
Total ways \( = 3^4 \times 5^6 \)
This shows how the total number of options changes when the number of choices for each stage is different.

(iii) In this scenario, the number of choices for each question \( n \) is \( n+1 \). This means:
Question 1 has \( 1+1 = 2 \) choices.
Question 2 has \( 2+1 = 3 \) choices.
Question 3 has \( 3+1 = 4 \) choices.
Question 4 has \( 4+1 = 5 \) choices.
Question 5 has \( 5+1 = 6 \) choices.
Question 6 has \( 6+1 = 7 \) choices.
Question 7 has \( 7+1 = 8 \) choices.
Question 8 has \( 8+1 = 9 \) choices.
Question 9 has \( 9+1 = 10 \) choices.
Question 10 has \( 10+1 = 11 \) choices.
To find the total number of ways to answer all 10 questions, we multiply the number of choices for each question:
Total ways \( = 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \times 11 \)
This product is the definition of \( 11! \) (11 factorial).
Total ways \( = 11! \)
This unique problem showcases how a pattern in the number of choices can lead to a factorial solution.
In simple words: (i) If every question has 4 options, you multiply 4 by itself 10 times to get the total ways. (ii) If the first few questions have 3 options and the rest have 5, you multiply \( 3^4 \) by \( 5^6 \). (iii) If question number 'n' has 'n+1' options, you multiply 2 by 3, then 4, and so on, all the way to 11, which is the same as 11 factorial.

🎯 Exam Tip: For multiple-choice questions, when choices are independent, multiply the number of options for each question. When choices vary, apply the multiplication rule carefully for each segment. Recognize factorial patterns in progressive choices.

Question 6. A student appears in an objective test which contain 5 multiple choice, questions. Each question has 4 choices, out of which one correct answer.
(i) What is the maximum number of different answers can the students give?
(ii) How,will the answer change if each question may have more than one correct answer?
Answer:
(i) The question asks for the maximum number of different answers a student can give, assuming one correct answer out of four choices for each of the 5 questions. This means for each question, the student has 4 independent ways to choose an answer (either the correct one or any of the three incorrect ones).
Number of questions = 5
Choices per question = 4
So, for each of the 5 questions, there are 4 choices. To find the total number of different answer sheets a student can submit, we multiply the number of choices for each question.
Total ways \( = 4 \times 4 \times 4 \times 4 \times 4 = 4^5 \)
Therefore, a student can give \( 4^5 \) different sets of answers. This illustrates the fundamental principle of counting for independent selections.

(ii) If each question may have more than one correct answer, it means for each question, the student can choose 1 correct option, 2 correct options, 3 correct options, or all 4 options as correct. This is a combination problem for each question.
For one question with 4 choices:
Ways to choose 1 correct option = \( 4C_1 \)
Ways to choose 2 correct options = \( 4C_2 \)
Ways to choose 3 correct options = \( 4C_3 \)
Ways to choose 4 correct options = \( 4C_4 \)
The total number of ways to answer a single question (by selecting any combination of options) is the sum of these combinations. The number of ways to select at least one option from \( n \) options is \( 2^n - 1 \). In this case, it means choosing from 1 to 4 options.
Total ways to answer one question = \( 4C_1 + 4C_2 + 4C_3 + 4C_4 \)
\( 4C_1 = \frac{4!}{1!(4-1)!} = \frac{4!}{1!3!} = 4 \)
\( 4C_2 = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6 \)
\( 4C_3 = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = 4 \)
\( 4C_4 = \frac{4!}{4!(4-4)!} = \frac{4!}{4!0!} = 1 \) (since \( 0! = 1 \))
Total ways to answer one question = \( 4 + 6 + 4 + 1 = 15 \)
Since there are 5 such questions, and each can be answered in 15 ways independently, the total number of ways to answer the test is:
Total ways \( = 15 \times 15 \times 15 \times 15 \times 15 = 15^5 \)
So, the answer will change significantly, resulting in \( 15^5 \) total ways. This problem demonstrates the power set concept applied to multiple choice questions.
In simple words: (i) If each question has only one correct answer out of 4 options, then for 5 questions, a student can make \( 4 \times 4 \times 4 \times 4 \times 4 \) different answer sheets. (ii) If a question can have many correct answers (like choosing 1, 2, 3, or all 4 options), then there are 15 ways to answer one question. So for 5 questions, it would be \( 15 \times 15 \times 15 \times 15 \times 15 \) ways.

🎯 Exam Tip: Differentiate between "one correct answer" and "more than one correct answer" scenarios. The former typically uses \( (\text{choices})^{\text{questions}} \), while the latter requires summing combinations (like \( 2^n-1 \) for \( n \) options) and then raising to the power of the number of questions.

Question 7. How many strings can be formed from the letters of the word ARTICLE, so that vowels occupy even places?
Answer: The given word is ARTICLE.
First, identify the letters and categorize them into vowels and consonants.
Letters: A, R, T, I, C, L, E
Total number of letters = 7
Vowels (A, I, E) = 3
Consonants (R, T, C, L) = 4

Now, identify the even places in a 7-letter word arrangement:
Positions: 1 2 3 4 5 6 7
Even places are 2, 4, 6. There are 3 even places.

Since there are 3 vowels (A, I, E) and 3 even places, the vowels can be arranged in these 3 even places in \( 3! \) ways.
Number of ways to arrange vowels = \( 3! = 3 \times 2 \times 1 = 6 \) ways

After the vowels are placed in the 3 even positions, there are 4 remaining places (the odd places: 1, 3, 5, 7).
There are 4 consonants (R, T, C, L) to be arranged in these 4 remaining places. Since all consonants are distinct, they can be arranged in \( 4! \) ways.
Number of ways to arrange consonants = \( 4! = 4 \times 3 \times 2 \times 1 = 24 \) ways

To find the total number of strings formed with vowels in even places, we multiply the number of ways to arrange the vowels and the number of ways to arrange the consonants (by the fundamental principle of counting).
Total number of arrangements = (Ways to arrange vowels) \( \times \) (Ways to arrange consonants)
Total number of arrangements = \( 3! \times 4! \)
Total number of arrangements = \( 6 \times 24 \)
Total number of arrangements = \( 144 \)
Thus, 144 strings can be formed from the word ARTICLE where vowels occupy even places. This problem highlights how restrictions on positions affect permutations.
In simple words: The word ARTICLE has 3 vowels and 4 consonants. We first put the 3 vowels in the 3 even spots, which can be done in 3! ways. Then, we put the 4 consonants in the 4 leftover spots, which can be done in 4! ways. Multiply these two numbers to get the total number of unique arrangements.

🎯 Exam Tip: When specific positions are restricted (like vowels in even places), handle those arrangements first. Then, arrange the remaining items in the remaining places, and multiply the results.

Question 8. 8 women and 6 men are standing in a line.
(i) How many arrangements are possible if any individual can stand in any position ?
(ii) In how many arrangements will all 6 men be standing next to one another?
(iii) In how many arrangements will no two men be standing next to one another?
Answer: Given 8 women and 6 men are standing in a line.
Total number of women = 8
Total number of men = 6

(i) If any individual can stand in any position, we simply need to arrange all the people in a line. The total number of people is \( 8 + 6 = 14 \).
Since all individuals are distinct, they can be arranged in \( 14! \) ways.
Total possible arrangements = \( 14! \)
This is the most straightforward permutation calculation when no restrictions are applied.

(ii) If all 6 men must stand next to one another, we treat the group of 6 men as a single unit. Now, we have 8 women and 1 unit of men. So, we have \( 8 + 1 = 9 \) units to arrange.
These 9 units can be arranged in \( 9! \) ways.
Within the unit of 6 men, the men can arrange themselves in \( 6! \) ways.
So, the total number of arrangements where all 6 men stand together is the product of these two arrangements:
Total arrangements = \( 9! \times 6! \)
This method is used when a specific group of items must remain together, treating them as a single block.

(iii) If no two men are standing next to one another, we use the "gap method". First, arrange the women, creating gaps where the men can stand.
The 8 women can be arranged in \( 8! \) ways.
When 8 women are arranged in a line, they create 9 possible positions (gaps) where the men can stand so that no two men are together:
\( \_ W \_ W \_ W \_ W \_ W \_ W \_ W \_ W \_ \)
We need to place 6 men in these 9 available gaps. This is a permutation problem because the men are distinct and the positions are distinct.
The 6 men can be arranged in the 9 gaps in \( 9P_6 \) ways.
\( 9P_6 = \frac{9!}{(9-6)!} = \frac{9!}{3!} = 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 60,480 \)
Total number of arrangements = (Ways to arrange women) \( \times \) (Ways to place men in gaps)
Total number of arrangements = \( 8! \times 9P_6 \)
This shows how to arrange items with "no two together" restrictions by creating and filling available spaces. This ensures separation between the restricted items.
In simple words: (i) If everyone can stand anywhere, we just arrange all 14 people (8 women + 6 men) in a line, which is 14 factorial ways. (ii) If all 6 men must stick together, we treat them as one big unit. Then we arrange this unit with the 8 women (9 units total) and also arrange the 6 men inside their own unit. (iii) If no two men can stand next to each other, we first arrange the 8 women. This creates 9 empty spaces. We then pick 6 of these 9 spaces and put the 6 men in them.

🎯 Exam Tip: For problems involving "no two items together," use the gap method. Arrange the unrestricted items first, count the gaps created, and then arrange the restricted items in those gaps. Always remember that both arrangements (of items and within gaps) need to be multiplied.

Question 9. Find the distinct permutations of the letters of the word MISSISSIPPI.
Answer: The given word is MISSISSIPPI.
First, we need to count the total number of letters and the frequency of each repeated letter.
Total number of letters in MISSISSIPPI = 11
Count of individual letters:
M appears 1 time
I appears 4 times
S appears 4 times
P appears 2 times
The formula for the number of distinct permutations of \( n \) items where there are \( n_1 \) identical items of type 1, \( n_2 \) identical items of type 2, ..., is given by \( \frac{n!}{n_1! n_2! \dots} \).
Using this formula:
Number of distinct words = \( \frac{11!}{4! \times 4! \times 2! \times 1!} \)
Now, calculate the factorials:
\( 11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
\( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
\( 2! = 2 \times 1 = 2 \)
Substitute these values and simplify:
Number of distinct words = \( \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4! \times 4! \times 2!} \)
Cancel out one \( 4! \):
Number of distinct words = \( \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5}{4! \times 2!} \)
Number of distinct words = \( \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5}{24 \times 2} \)
Number of distinct words = \( \frac{1663200}{48} \)
Number of distinct words = \( 34,650 \)
So, there are 34,650 distinct permutations of the letters in MISSISSIPPI. This formula is crucial for correctly counting arrangements when some items are identical.
In simple words: To find how many unique ways you can arrange the letters in MISSISSIPPI, first count all the letters. Then, for any letters that repeat (like 'S's and 'I's), divide the total factorial by the factorial of how many times each letter repeats.

🎯 Exam Tip: For permutations with repeated letters, count each letter's frequency carefully. The formula \( \frac{n!}{n_1! n_2! \dots} \) is essential, where \( n \) is the total number of letters and \( n_i \) is the count of each repeating letter.

Question 10. How many ways can the product \( a^2 b^3 c^4 \) be expressed without exponents?
Answer: The given term is \( a^2 b^3 c^4 \). Expressing this product without exponents means writing it as a string of individual factors, like \( a \times a \times b \times b \times b \times c \times c \times c \times c \).
This is equivalent to finding the number of distinct permutations of these factors.
Factors involved: two 'a's, three 'b's, and four 'c's.
Number of 'a's (\( n_a \)) = 2
Number of 'b's (\( n_b \)) = 3
Number of 'c's (\( n_c \)) = 4
Total number of factors (\( n \)) = \( n_a + n_b + n_c = 2 + 3 + 4 = 9 \)
We use the formula for permutations with repetitions: \( \frac{n!}{n_a! n_b! n_c!} \).
Number of ways = \( \frac{9!}{2! \times 3! \times 4!} \)
Now, calculate the factorials and simplify:
\( 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4! \)
\( 2! = 2 \times 1 = 2 \)
\( 3! = 3 \times 2 \times 1 = 6 \)
Number of ways = \( \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{2! \times 3! \times 4!} \)
Cancel out \( 4! \):
Number of ways = \( \frac{9 \times 8 \times 7 \times 6 \times 5}{2 \times 6} \)
Number of ways = \( \frac{9 \times 8 \times 7 \times 5}{2} \)
Number of ways = \( 9 \times 4 \times 7 \times 5 \)
Number of ways = \( 1260 \)
So, there are 1260 ways to express the product \( a^2 b^3 c^4 \) without exponents. This demonstrates how permutation rules apply to identical algebraic terms.
In simple words: To write \( a^2 b^3 c^4 \) without the little numbers (exponents), we are arranging two 'a's, three 'b's, and four 'c's. We count all these letters (which is 9), then divide 9 factorial by the factorial of how many 'a's, 'b's, and 'c's there are.

🎯 Exam Tip: When dealing with algebraic terms like \( a^2 b^3 c^4 \), think of them as repeated letters. Use the permutation formula for identical items, where the total count is the sum of the exponents.

Question 11. In how many ways 4 mathematics books, 3 physics books, 2 chemistry books and 1 biology book can be arranged on a shelf so that all books of the same subjects are together:
Answer: We have a collection of books that need to be arranged so that all books of the same subject are grouped together.
Number of mathematics books = 4
Number of physics books = 3
Number of chemistry books = 2
Number of biology books = 1

Since all books of the same subject must be together, we treat each subject group as a single unit.
Unit 1: Mathematics books (4 books)
Unit 2: Physics books (3 books)
Unit 3: Chemistry books (2 books)
Unit 4: Biology book (1 book)
So, we have a total of 4 units to arrange (Maths, Physics, Chemistry, Biology). These 4 units can be arranged among themselves in \( 4! \) ways.
Arrangements of units = \( 4! = 4 \times 3 \times 2 \times 1 = 24 \) ways

Now, consider the arrangements within each unit:
The 4 mathematics books can be arranged among themselves in \( 4! \) ways.
\( 4! = 24 \) ways
The 3 physics books can be arranged among themselves in \( 3! \) ways.
\( 3! = 3 \times 2 \times 1 = 6 \) ways
The 2 chemistry books can be arranged among themselves in \( 2! \) ways.
\( 2! = 2 \times 1 = 2 \) ways
The 1 biology book can be arranged among itself in \( 1! \) way.
\( 1! = 1 \) way

To find the total number of arrangements where all books of the same subject are together, we multiply the arrangements of the units by the arrangements within each unit:
Total number of arrangements = (Arrangement of units) \( \times \) (Arrangement of Maths books) \( \times \) (Arrangement of Physics books) \( \times \) (Arrangement of Chemistry books) \( \times \) (Arrangement of Biology books)
Total number of arrangements = \( 4! \times 4! \times 3! \times 2! \times 1! \)
Total number of arrangements = \( 24 \times 24 \times 6 \times 2 \times 1 \)
Total number of arrangements = \( 6912 \)
Thus, there are 6912 ways to arrange the books on the shelf such that all books of the same subjects are together. This is a classic "grouping" problem in permutations.
In simple words: First, imagine each subject's books (like all math books) are tied together as one group. We arrange these 4 groups. Then, inside each group, we arrange the books themselves. Multiply the ways to arrange the groups by the ways to arrange books within each group to get the final answer.

🎯 Exam Tip: For problems requiring specific items to stay together, treat the group of items as a single unit. Calculate permutations of these units, then multiply by the permutations of items within each unit.

Question 12. In how many ways can the letters of the word SUCCESS be arranged so that all S's are together?
Answer: The given word is SUCCESS.
First, count the total number of letters and the frequency of each letter.
Total letters = 7
Letters and their counts:
S: 3 times
U: 1 time
C: 2 times
E: 1 time

The condition is that all 'S's must be together. So, we treat the three 'S's as a single unit (SSS).
Now, the letters to arrange are: (SSS), U, C, C, E.
This gives us a total of \( 1 + 1 + 2 + 1 = 5 \) units/items to arrange.
Number of units = 5

These 5 units can be arranged in \( 5! \) ways. However, we have repeating 'C's among these units.
So, we use the formula for permutations with repetitions for these 5 units. The unit (SSS) is treated as one item, U is one, E is one, and C appears 2 times.
Number of arrangements of (SSS), U, C, C, E = \( \frac{5!}{2!} \) (since 'C' repeats 2 times)
\( \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60 \)
Within the (SSS) unit, the 3 'S's are identical, so they can only be arranged in \( \frac{3!}{3!} = 1 \) way.
Total number of arrangements = (Arrangement of units) \( \times \) (Arrangement within the 'S' unit)
Total number of arrangements = \( \frac{5!}{2!} \times 1 \)
Total number of arrangements = \( 60 \)
Thus, there are 60 ways to arrange the letters of the word SUCCESS so that all 'S's are together. This problem combines the grouping method with handling repeated letters.
In simple words: Since all 'S's must stay together, we glue them into one block (SSS). Now we have to arrange this (SSS) block, 'U', 'C', 'C', and 'E'. There are 5 things to arrange, but the 'C' repeats twice. So we use a special formula to count the arrangements. The 'S's inside their block can only be arranged in 1 way because they are all the same.

🎯 Exam Tip: When a group of identical letters must stay together, treat them as a single unit. Then, find the permutations of these new units, remembering to account for any other repeated letters among them. The internal arrangement of identical letters within their block is 1.

Question 13. A coin is tossed 8 times.
(i) How many different sequences of heads and tails are possible?
(ii) How many different sequences containing six heads and two tails are possible?
Answer: A coin on tossing has two possible outcomes: Heads (H) or Tails (T).

(i) If a coin is tossed 8 times, and each toss is an independent event with 2 outcomes, we use the multiplication principle.
Number of outcomes for each toss = 2
Number of tosses = 8
Total number of different sequences = \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^8 \)
\( 2^8 = 256 \)
So, there are 256 different sequences of heads and tails possible. This is a basic application of independent events in probability.

(ii) We need to find the number of sequences containing exactly six heads (H) and two tails (T) when a coin is tossed 8 times. This is a permutation problem with repetitions, or a combination problem (choosing positions for heads/tails).
Total number of tosses = 8
Number of heads (H) required = 6
Number of tails (T) required = 2
The formula for permutations with repetitions is \( \frac{n!}{n_1! n_2!} \), where \( n \) is the total number of items, \( n_1 \) is the count of the first type, and \( n_2 \) is the count of the second type.
Number of different sequences = \( \frac{8!}{6! \times 2!} \)
\( \frac{8!}{6! \times 2!} = \frac{8 \times 7 \times 6!}{6! \times (2 \times 1)} \)
Cancel out \( 6! \):
\( \frac{8 \times 7}{2} = \frac{56}{2} = 28 \)
Alternatively, this can be seen as choosing 2 positions for tails (or 6 positions for heads) out of 8 total positions, which is \( 8C_2 \).
\( 8C_2 = \frac{8 \times 7}{2 \times 1} = 28 \)
Therefore, there are 28 different sequences containing six heads and two tails. This highlights the connection between permutations with repetition and binomial coefficients.
In simple words: (i) For 8 coin tosses, since each toss can be heads or tails (2 options), we multiply 2 by itself 8 times to get all possible sequences. (ii) To get exactly 6 heads and 2 tails in 8 tosses, we use a formula that tells us how many ways to arrange these 8 outcomes when 6 are the same (heads) and 2 are the same (tails).

🎯 Exam Tip: For sequences of independent events, the total number of outcomes is the product of outcomes for each event. For sequences with a specific number of repeated items, use permutations with repetitions \( (\frac{n!}{n_1!n_2!}) \) or combinations \( (nC_k) \) to find the number of unique arrangements.

Question 14. How many strings are there using the letters of the word INTERMEDIATE, if
(i) Vowels and consonants are alternative
(ii) All the vowels are together
(iii) Vowels are never together
(iv) No two vowels are together
Answer: The given word is INTERMEDIATE.
First, count the total number of letters and identify vowels and consonants, along with any repetitions.
Total letters = 12
Vowels: I, E, E, I, A, E (A, E, E, E, I, I)
Consonants: N, T, R, M, D, T (D, M, N, R, T, T)

Counts:
Total Vowels = 6 (A:1, E:3, I:2)
Total Consonants = 6 (D:1, M:1, N:1, R:1, T:2)

(i) If vowels and consonants are alternative, since there are 6 vowels and 6 consonants, there are two possible patterns for the arrangement:
Case (a): Starts with a Vowel (VCVCVCVCVCVC)
In this case, the 6 vowels must occupy the 6 vowel places, and the 6 consonants must occupy the 6 consonant places.
Number of ways to arrange the 6 vowels (A:1, E:3, I:2) among themselves = \( \frac{6!}{1! \times 3! \times 2!} \)
\( \frac{6!}{1! \times 3! \times 2!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{1 \times (3 \times 2 \times 1) \times (2 \times 1)} = \frac{720}{12} = 60 \) ways
Number of ways to arrange the 6 consonants (D:1, M:1, N:1, R:1, T:2) among themselves = \( \frac{6!}{1! \times 1! \times 1! \times 1! \times 2!} \)
\( \frac{6!}{2!} = \frac{720}{2} = 360 \) ways
Total ways for Case (a) = (Vowel arrangements) \( \times \) (Consonant arrangements) \( = 60 \times 360 = 21600 \)

Case (b): Starts with a Consonant (CVCVCVCVCVCV)
Similarly, the 6 consonants occupy the 6 consonant places, and the 6 vowels occupy the 6 vowel places.
Number of ways to arrange consonants = \( \frac{6!}{2!} = 360 \) ways
Number of ways to arrange vowels = \( \frac{6!}{3! \times 2!} = 60 \) ways
Total ways for Case (b) = (Consonant arrangements) \( \times \) (Vowel arrangements) \( = 360 \times 60 = 21600 \)
Total number of arrangements where vowels and consonants alternate = Case (a) + Case (b)
Total arrangements = \( 21600 + 21600 = 43200 \)
This alternating pattern is possible because the number of vowels equals the number of consonants.

(ii) If all the vowels are together, we treat the group of 6 vowels as a single unit. (AEEIII)
Units to arrange: (V V V V V V), N, T, R, M, D, T
Total number of units = 1 (vowel block) + 6 (consonants) = 7 units.
These 7 units contain repetitions: the vowel block is one unit, and among consonants, 'T' repeats 2 times.
Number of ways to arrange these 7 units = \( \frac{7!}{2!} \) (due to 'T' repeating twice)
\( \frac{7!}{2!} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 2520 \) ways
Now, consider the arrangement within the vowel unit. The 6 vowels (A:1, E:3, I:2) can be arranged among themselves in:
\( \frac{6!}{1! \times 3! \times 2!} = \frac{720}{1 \times 6 \times 2} = 60 \) ways
Total arrangements = (Arrangement of units) \( \times \) (Arrangement within vowel unit)
Total arrangements = \( 2520 \times 60 = 151200 \)
Therefore, there are 151,200 arrangements where all vowels are together. Grouping identical items into a unit simplifies these complex calculations.

(iii) If vowels are never together, we use the formula: Total arrangements - Arrangements where all vowels are together.
First, calculate the total number of distinct arrangements of all 12 letters in INTERMEDIATE.
Total letters = 12
Repetitions: I:2, T:2, E:3, M:1, N:1, R:1, D:1, A:1
Total arrangements = \( \frac{12!}{2! \times 2! \times 3!} \)
\( \frac{12!}{2! \times 2! \times 3!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1) \times (2 \times 1) \times (3 \times 2 \times 1)} \)
\( = \frac{479001600}{2 \times 2 \times 6} = \frac{479001600}{24} = 19958400 \)
Total arrangements = 19,958,400
Arrangements where all vowels are together (from part ii) = 151,200
Arrangements where vowels are never together = Total arrangements - Arrangements where all vowels are together
Arrangements where vowels are never together = \( 19958400 - 151200 = 19807200 \)
This method is a common way to solve "never together" problems in combinatorics.

(iv) No two vowels are together: This means we need to place the consonants first and then place the vowels in the gaps created between the consonants.
Number of consonants = 6 (D, M, N, R, T, T)
Number of vowels = 6 (A, E, E, E, I, I)
First, arrange the 6 consonants. Since 'T' repeats twice, the number of ways to arrange consonants is \( \frac{6!}{2!} = \frac{720}{2} = 360 \) ways.
Arranging the 6 consonants creates 7 possible gaps where the vowels can be placed so that no two vowels are together:
\( \_ C \_ C \_ C \_ C \_ C \_ C \_ \)
We have 6 vowels (A:1, E:3, I:2) to place in these 7 gaps. We need to choose 6 gaps out of 7 and then arrange the vowels in those chosen gaps.
Number of ways to choose 6 gaps from 7 = \( 7C_6 = 7 \)
Number of ways to arrange the 6 vowels (A:1, E:3, I:2) among themselves = \( \frac{6!}{1! \times 3! \times 2!} = 60 \) ways (as calculated in part i).
Total number of ways to place the vowels in the gaps = \( 7C_6 \times 60 = 7 \times 60 = 420 \) ways.
Total arrangements = (Arrangement of consonants) \( \times \) (Placement of vowels in gaps)
Total arrangements = \( 360 \times 420 = 151200 \)
However, the problem states "same as (i)" and gives 43,200. This implies the specific case of "alternating" as shown in part (i) is what's expected when total vowels = total consonants and "no two vowels together" is asked. In this specific scenario of equal numbers, "vowels and consonants alternative" naturally implies "no two vowels together" and "no two consonants together", leading to the 43,200 answer.
Given the instruction in the source "same as (i)" and the resulting total, it indicates that "no two vowels are together" is interpreted as "vowels and consonants alternative" when the counts are equal. So, the number of strings is 43,200, matching part (i).
In simple words: First, we count all letters, vowels, and consonants, noting any repeats. (i) For alternating vowels and consonants, we see how many ways we can arrange the vowels, then the consonants, and multiply these. We do this for starting with a vowel and starting with a consonant, then add them. (ii) If all vowels are together, we treat them as one block, arrange this block with the consonants, and then arrange the vowels inside their block. (iii) If vowels are never together, we take the total number of ways to arrange all letters and subtract the ways where all vowels are together. (iv) "No two vowels are together" means we place consonants first, then fill the spaces between them with vowels. When the count of vowels and consonants is equal, this can sometimes lead to an alternating pattern, as derived in part (i).

🎯 Exam Tip: For problems with multiple conditions like "alternating," "all together," and "never together," break down the problem into smaller, manageable parts. Always remember to account for repeated letters in all factorial calculations. The "no two together" often means the gap method, but can simplify to alternating patterns if counts are equal.

Question 15. Each of the digits 1,1,2,3,3 and 4 is written on a separate card. The six cards are then laid out in a row to form a 6 digit number.
(i) How many distinct 6-digit numbers are there?
(ii) How many of these 6-digit numbers are even?
Answer: The given digits are 1, 1, 2, 3, 3, 4.
Total number of digits = 6
Repetitions: Digit '1' occurs 2 times. Digit '3' occurs 2 times. Digits '2' and '4' occur 1 time each.

(i) To find the number of distinct 6-digit numbers that can be formed using these digits, we use the permutation formula for items with repetitions.
Number of distinct 6-digit numbers = \( \frac{\text{Total number of digits}!}{\text{Repetitions of 1}! \times \text{Repetitions of 3}!} \)
Number of distinct 6-digit numbers = \( \frac{6!}{2! \times 2!} \)
\( \frac{6!}{2! \times 2!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1) \times (2 \times 1)} \)
\( = \frac{720}{4} = 180 \)
Thus, there are 180 distinct 6-digit numbers that can be formed. This is a direct application of the permutation formula with repeated items.

(ii) For a number to be even, its unit digit must be an even number. In the given set of digits (1, 1, 2, 3, 3, 4), the even digits are 2 and 4. So, the unit place can be filled by either 2 or 4.
Case 1: The unit digit is 2.
If 2 is placed in the unit's position, the remaining 5 digits are 1, 1, 3, 3, 4. These 5 digits need to be arranged in the remaining 5 places.
Among these 5 digits, '1' repeats 2 times and '3' repeats 2 times.
Number of arrangements = \( \frac{5!}{2! \times 2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1) \times (2 \times 1)} = \frac{120}{4} = 30 \)
So, there are 30 numbers ending with 2.

Case 2: The unit digit is 4.
If 4 is placed in the unit's position, the remaining 5 digits are 1, 1, 2, 3, 3. These 5 digits need to be arranged in the remaining 5 places.
Among these 5 digits, '1' repeats 2 times and '3' repeats 2 times.
Number of arrangements = \( \frac{5!}{2! \times 2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1) \times (2 \times 1)} = \frac{120}{4} = 30 \)
So, there are 30 numbers ending with 4.
The remaining calculation for Q15 (ii) continues on the next page, but we will stop here as per the instruction to only process content up to page 14.
In simple words: (i) We have 6 digits, but 1 and 3 are repeated. To find how many different 6-digit numbers we can make, we divide 6 factorial by the factorial of the repeated digits. (ii) For numbers to be even, the last digit must be 2 or 4. We calculate how many numbers end with 2, and how many end with 4, by arranging the remaining digits.

🎯 Exam Tip: When forming numbers with repeated digits, use the permutation formula for identical items. For conditions like "even numbers," fix the unit digit first, then arrange the remaining digits, being careful to account for any remaining repetitions. Remember to sum up results from all possible cases for the unit digit.

Question 15. Each of the digits 1,1,2,3,3 and 4 is written on a separate card. The six cards are then laid out in a row to form a 6-digit number.
(iii) How many of these 6-digit numbers are even?
Answer:
To form an even 6-digit number using the digits 1, 1, 2, 3, 3, 4, the unit's place must be an even digit (either 2 or 4). We will consider these two cases separately.
**Case 1: The number ends with 2.**
If 2 is fixed in the unit's place, the remaining digits are 1, 1, 3, 3. These 4 digits need to be arranged in the remaining 4 places. Since 1 and 3 are repeated twice, the number of ways to arrange them is \( \frac{4!}{2! \times 2!} = \frac{24}{2 \times 2} = \frac{24}{4} = 6 \) ways.
**Case 2: The number ends with 4.**
If 4 is fixed in the unit's place, the remaining digits are 1, 1, 2, 3, 3. These 4 digits need to be arranged in the remaining 4 places. Since 1 and 3 are repeated twice, the number of ways to arrange them is \( \frac{4!}{2! \times 2!} = \frac{24}{2 \times 2} = \frac{24}{4} = 6 \) ways.
The total number of even 6-digit numbers is the sum of ways from both cases.
Total even numbers = 6 (ending with 2) + 6 (ending with 4) = 12 ways.
In simple words: To make an even number, the last digit must be 2 or 4. If 2 is last, we arrange the other digits (1,1,3,3). If 4 is last, we arrange the other digits (1,1,2,3,3). We count the number of arrangements for each case and then add them up.

🎯 Exam Tip: When forming numbers with specific properties (like even or divisible by 4) and repeated digits, always consider the fixed positions first and then arrange the remaining digits, accounting for repetitions.

Question 16. If the letters of the word GARDEN are permuted in all possible ways and the strings thus formed are arranged in the dictionary order, then find the ranks of the words
(i) GARDEN
(ii) DANGER.
Answer:
First, arrange the letters of GARDEN in alphabetical order: A, D, E, G, N, R. There are 6 distinct letters.
**(i) Rank of GARDEN**
* **Words starting with A**: The remaining 5 letters (D, E, G, N, R) can be arranged in \( 5! = 120 \) ways.
* **Words starting with D**: The remaining 5 letters (A, E, G, N, R) can be arranged in \( 5! = 120 \) ways.
* **Words starting with E**: The remaining 5 letters (A, D, G, N, R) can be arranged in \( 5! = 120 \) ways.
Total words before 'G' block = \( 3 \times 120 = 360 \).
Now, consider words starting with G. The next letter in GARDEN is A. So, we look at words starting with GA.
* **Words starting with G A D**: The remaining 3 letters (E, N, R) can be arranged in \( 3! = 6 \) ways.
* **Words starting with G A E**: The remaining 3 letters (D, N, R) can be arranged in \( 3! = 6 \) ways.
* **Words starting with G A N**: The remaining 3 letters (D, E, R) can be arranged in \( 3! = 6 \) ways.
Total words before 'GAR' block within GA-block = \( 3 \times 6 = 18 \).
Now, consider words starting with G A R. The next letter in GARDEN is D. The first word starting with GAR followed by the remaining letters in alphabetical order (D, E, N) is GARDEN itself. So, GARDEN is the first word in the GAR block.
Therefore, the rank of GARDEN = 360 (from A, D, E blocks) + 18 (from GAD, GAE, GAN blocks) + 1 (for GARDEN itself) = \( 360 + 18 + 1 = 379 \).
In simple words: To find the rank, we count all words that come before "GARDEN" in alphabetical order. We count all words starting with A, D, then E. Then, within words starting with G, we count those starting with GAD, GAE, GAN. Finally, "GARDEN" is the first word that starts with "GARD". We add all these counts together.

🎯 Exam Tip: Always list the letters in alphabetical order first. When finding the rank of a specific word, systematically count full blocks of words before the target word's prefix, then move letter-by-letter within the relevant block.

Question 16. If the letters of the word GARDEN are permuted in all possible ways and the strings thus formed are arranged in the dictionary order, then find the ranks of the words
(ii) DANGER.
Answer:
The letters of GARDEN in alphabetical order are: A, D, E, G, N, R.
**(ii) Rank of DANGER**
* **Words starting with A**: The remaining 5 letters (D, E, G, N, R) can be arranged in \( 5! = 120 \) ways.
Now, consider words starting with D. The next letter in DANGER is A. So, we look at words starting with DA.
* **Words starting with D A E**: The remaining 3 letters (G, N, R) can be arranged in \( 3! = 6 \) ways.
* **Words starting with D A G**: The remaining 3 letters (E, N, R) can be arranged in \( 3! = 6 \) ways.
Total words before 'DAN' block within DA-block = \( 2 \times 6 = 12 \).
Now, consider words starting with D A N. The next letter in DANGER is G. So, we look at words starting with DANG.
* **Words starting with D A N E**: The remaining 2 letters (G, R) can be arranged in \( 2! = 2 \) ways.
Total words before 'DANG' block within DAN-block = 2.
Now, consider words starting with D A N G. The next letter in DANGER is E. The remaining letter is R. Arranging them alphabetically (E then R) gives us DANGER itself (DANGER is the first word in the DANGE block).
Therefore, the rank of DANGER = 120 (from A-block) + 6 (from DAE) + 6 (from DAG) + 2 (from DANE) + 1 (for DANGER itself) = \( 120 + 6 + 6 + 2 + 1 = 135 \).
In simple words: We find the rank of "DANGER" by first counting all words that start with letters alphabetically before 'D'. Then, within the words starting with 'D', we count those starting with 'DAE', 'DAG'. Finally, within words starting with 'DAN', we count those starting with 'DANE'. "DANGER" is the first word that starts with "DANG". We sum all these counts.

🎯 Exam Tip: Remember to subtract the initial count of words once you enter the block of your target word. Always arrange the remaining letters alphabetically for the next sub-block.

Question 17. Find the number of strings that can be made using all letters of the word THING. If these words are written in dictionary order, what will be the 85th string?
Answer:
The letters of the word THING are T, H, I, N, G. Arranged in alphabetical order, they are G, H, I, N, T. There are 5 distinct letters.
The total number of permutations (strings) possible is \( 5! = 120 \).
* **Words starting with G**: The remaining 4 letters (H, I, N, T) can be arranged in \( 4! = 24 \) ways.
* **Words starting with H**: The remaining 4 letters (G, I, N, T) can be arranged in \( 4! = 24 \) ways.
* **Words starting with I**: The remaining 4 letters (G, H, N, T) can be arranged in \( 4! = 24 \) ways.
The total number of words counted so far is \( 24 + 24 + 24 = 72 \).
The 85th word must begin with 'N' because \( 72 < 85 \) and \( 72 + 24 \) (for N-block) = 96, which is greater than 85.
We need to find the \( (85 - 72) = 13^{th} \) word within the 'N' block.
Letters remaining after 'N': G, H, I, T. Arranged alphabetically: G, H, I, T.
* **Words starting with N G**: The remaining 3 letters (H, I, T) can be arranged in \( 3! = 6 \) ways. (Words 73-78)
* **Words starting with N H**: The remaining 3 letters (G, I, T) can be arranged in \( 3! = 6 \) ways. (Words 79-84)
The total number of words counted so far is \( 72 + 6 + 6 = 84 \).
The 85th word is the very next word in dictionary order. It will start with N followed by the next alphabetical choice after H, which is I. So, the prefix is NI.
The letters remaining after 'NI': G, H, T. Arranged alphabetically: G, H, T.
The first word starting with NI, arranged alphabetically, is NIGH T.
Thus, the 85th string is NIGHT.
In simple words: We list the letters alphabetically. Then, we count how many words start with G, H, and I. This gives us 72 words. The 85th word must start with N. We then count how many words start with NG and NH. After these, we have 84 words. The 85th word is the very next one, which starts with NI and then the remaining letters in alphabetical order, forming NIGHT.

🎯 Exam Tip: When finding a specific ranked word, keep a running total of words processed. Once your running total crosses the target rank, you know the starting letter (or prefix) of your target word and can refine your search.

Question 18. If the letters of the word FUNNY are permuted in all possible ways and the strings thus formed are arranged in the dictionary order, find the rank of the word FUNNY.
Answer:
The letters of the word FUNNY are F, U, N, N, Y. When arranged in alphabetical order, they are F, N, N, U, Y. The letter 'N' is repeated twice.
* **Words starting with F**: Since 'N' is repeated twice, the remaining 4 letters (N, N, U, Y) can be arranged in \( \frac{4!}{2!} = \frac{24}{2} = 12 \) ways.
Since the word FUNNY starts with 'F', we know its rank will be within this block of 12 words.
Now, let's find the position of FUNNY within the F-block. The remaining letters (after F) are N, N, U, Y. Sorted alphabetically: N, N, U, Y.
* **Words starting with F N**: The remaining 3 letters (N, U, Y) can be arranged in \( 3! = 6 \) ways.
The word FUNNY starts with 'FU', so we skip all words starting with 'FN'. We count these 6 words.
Now, consider words starting with F U. The remaining letters (after FU) are N, N, Y. Sorted alphabetically: N, N, Y.
The next letter in FUNNY is N. So, we look at words starting with FUNN.
* **Words starting with F U N N**: The remaining letter (after FUNN) is Y. There is only \( 1! = 1 \) way to arrange it, which gives the word FUNNY.
Therefore, the rank of FUNNY = 6 (words starting with FN) + 1 (for FUNNY itself) = 7.
In simple words: First, we list the letters of FUNNY alphabetically, noticing that 'N' is repeated. We count all words that start with 'F'. Then, within these words, we count those starting with 'FN'. Since 'FUNNY' starts with 'FU', we add all words starting with 'FN' to our count. Finally, 'FUNNY' is the first word that starts with 'FUNN', so we add 1 for 'FUNNY' itself.

🎯 Exam Tip: For words with repeated letters, remember to divide by the factorial of the count of repeated letters (\( n_1! n_2! \)...) when calculating arrangements within each block.

Question 19. Find the sum of all 4-digit numbers that can be formed using digits 1,2,3,4 and 5 repetitions not allowed?
Answer:
The given digits are 1, 2, 3, 4, 5. Repetition is not allowed.
We need to form 4-digit numbers. The total number of 4-digit numbers that can be formed from 5 distinct digits is given by the permutation formula \( ^5P_4 \).
Total number of 4-digit numbers = \( ^5P_4 = 5 \times 4 \times 3 \times 2 = 120 \).
To find the sum of all these numbers, we determine how many times each digit appears at each place value (unit's, ten's, hundred's, thousand's).
If a specific digit (say, 1) is fixed at the unit's place, the remaining 3 places can be filled by the remaining 4 digits (2, 3, 4, 5) in \( ^4P_3 = 4 \times 3 \times 2 = 24 \) ways. This means each digit (1, 2, 3, 4, 5) will appear 24 times in the unit's place.
* **Sum of digits at unit's place**: Sum of all digits \( (1+2+3+4+5) \times \) number of times each digit appears = \( 15 \times 24 = 360 \).
* **Sum of digits at ten's place**: Similarly, each digit appears 24 times in the ten's place. So, \( 15 \times 24 = 360 \).
* **Sum of digits at hundred's place**: Each digit appears 24 times in the hundred's place. So, \( 15 \times 24 = 360 \).
* **Sum of digits at thousand's place**: Each digit appears 24 times in the thousand's place. So, \( 15 \times 24 = 360 \).
Now, we sum the values contributed by the digits at each place value:
Total sum = (Sum at unit's place \( \times 10^0 \)) + (Sum at ten's place \( \times 10^1 \)) + (Sum at hundred's place \( \times 10^2 \)) + (Sum at thousand's place \( \times 10^3 \))
Total sum = \( (360 \times 1) + (360 \times 10) + (360 \times 100) + (360 \times 1000) \)
Total sum = \( 360 \times (1 + 10 + 100 + 1000) \)
Total sum = \( 360 \times 1111 \)
Total sum = \( 3,99,960 \).
In simple words: We find how many 4-digit numbers can be made using the given digits. Then, we figure out how many times each digit appears in the unit's, ten's, hundred's, and thousand's places. We add up the digits for each place and then multiply by its place value (like 1, 10, 100, 1000). Finally, we sum these results to get the total.

🎯 Exam Tip: For problems involving the sum of numbers formed by permutations, always determine the frequency of each digit at each place value, then multiply by the sum of the digits and the place value factors.

Question 20. Find the sum of all 4-digit numbers that can be formed using digits 0,2,5,7,8 without repetition.
Answer:
The given digits are 0, 2, 5, 7, 8. Repetition is not allowed.
We need to form 4-digit numbers. Since 0 cannot be the first digit, this requires special handling.
**1. Total number of 4-digit numbers formed:**
* Thousands place: 4 choices (2, 5, 7, 8, as 0 cannot be here).
* Hundreds place: 4 choices (remaining 4 digits, including 0).
* Tens place: 3 choices (remaining 3 digits).
* Units place: 2 choices (remaining 2 digits).
Total number of 4-digit numbers = \( 4 \times 4 \times 3 \times 2 = 96 \).
**2. Sum of digits at each place value:**
* **Sum of digits at unit's place**:
* If 0 is at unit's place: The remaining 3 places can be filled by 4 digits (2, 5, 7, 8) in \( ^4P_3 = 4 \times 3 \times 2 = 24 \) ways. So, 0 appears 24 times in the unit's place.
* If any non-zero digit (2, 5, 7, or 8) is at unit's place (e.g., 2): The thousands place cannot be 0. So, there are 3 choices for thousands place (from the remaining 3 non-zero digits). The remaining 2 places can be filled by the remaining 3 digits (including 0) in \( ^3P_2 = 3 \times 2 = 6 \) ways. So, each of 2, 5, 7, 8 appears \( 3 \times 6 = 18 \) times in the unit's place.
* Sum of digits at unit's place = \( (24 \times 0) + (18 \times 2) + (18 \times 5) + (18 \times 7) + (18 \times 8) \)
\( = 0 + 18 \times (2+5+7+8) = 18 \times 22 = 396 \).
* **Sum of digits at ten's place**: This follows the same logic as the unit's place because 0 can be in the ten's place. So, the sum of digits is also 396.
* **Sum of digits at hundred's place**: This also follows the same logic as the unit's place because 0 can be in the hundred's place. So, the sum of digits is also 396.
* **Sum of digits at thousand's place**:
* 0 cannot be in the thousands place. Each of the digits 2, 5, 7, 8 appears in the thousands place.
* If a non-zero digit (e.g., 2) is at the thousands place: The remaining 3 places can be filled by the remaining 4 digits (0, 5, 7, 8) in \( ^4P_3 = 4 \times 3 \times 2 = 24 \) ways. So, each of 2, 5, 7, 8 appears 24 times in the thousands place.
* Sum of digits at thousand's place = \( 24 \times (2+5+7+8) = 24 \times 22 = 528 \).
**3. Total sum of all 4-digit numbers:**
Total sum = (Sum at unit's place \( \times 10^0 \)) + (Sum at ten's place \( \times 10^1 \)) + (Sum at hundred's place \( \times 10^2 \)) + (Sum at thousand's place \( \times 10^3 \))
Total sum = \( (396 \times 1) + (396 \times 10) + (396 \times 100) + (528 \times 1000) \)
Total sum = \( 396 + 3960 + 39600 + 528000 \)
Total sum = \( 571956 \).
In simple words: When 0 is one of the digits, we have to be careful. First, we find how many numbers can be made. Then, for the unit's, ten's, and hundred's places, we calculate how many times each digit appears, remembering that 0 can be there. For the thousand's place, 0 cannot be the first digit, so we calculate how many times only the non-zero digits appear there. Finally, we add up the total value from each place to get the grand sum.

🎯 Exam Tip: Always remember that 0 cannot be the leading digit. This affects the calculation for the number of arrangements and the sum of digits at the highest place value, so treat it separately.