Samacheer Kalvi Class 11 Business Maths Solutions Chapter 2 Algebra Exercise 2.3

Question 1. If \( nP_4 = 12(nP_2) \), find n.
Answer: We are given the equation \( nP_4 = 12(nP_2) \). We know that \( nP_r = \frac{n!}{(n-r)!} \). Using this formula, we can write the equation as:
\( \frac{n!}{(n-4)!} = 12 \times \frac{n!}{(n-2)!} \)
Now, we expand the factorials: \( n(n-1)(n-2)(n-3)(n-4)! = 12 \times n(n-1)(n-2)! \).
This simplifies to:
\( n(n-1)(n-2)(n-3) = 12n(n-1) \)
Since n must be greater than or equal to 4 (because of \( nP_4 \)), \( n \) and \( n-1 \) are not zero. We can cancel \( n(n-1) \) from both sides:
\( (n-2)(n-3) = 12 \)
We need to find two consecutive numbers that multiply to 12. We know that \( 4 \times 3 = 12 \).
So, we can set:
\( n-2 = 4 \)
\( \implies n = 4 + 2 \)
\( \implies n = 6 \)
Alternatively, \( n-3 = 3 \)
\( \implies n = 3 + 3 \)
\( \implies n = 6 \)
Therefore, the value of n is 6. This type of problem often tests your understanding of permutation definitions.
In simple words: We used the formula for permutations and then simplified the equation by canceling common terms. We found two consecutive numbers that multiply to 12, which helped us solve for n, giving us 6.

🎯 Exam Tip: Always remember the formula for permutations, \( nP_r = \frac{n!}{(n-r)!} \), and be careful when expanding factorials to simplify equations.

Question 2. In how many ways 5 boys and 3 girls can be seated in a row so that no two girls are together?
Answer: First, let's seat the 5 boys. The 5 boys can be seated among themselves in \( 5P_5 = 5! \) ways. This means there are 120 ways to arrange the boys.
When the 5 boys are seated, they create 6 possible places where the girls can sit so that no two girls are together. We can imagine the boys (B) creating spaces (X) like this:
X B X B X B X B X B X
There are 6 possible places (marked with X) where the 3 girls can be seated. Since the girls must not sit together, they have to occupy these separate spaces. We need to choose 3 of these 6 places and arrange the 3 girls in them.
The number of ways to seat the 3 girls in 6 available places is \( 6P_3 \).
\( 6P_3 = \frac{6!}{(6-3)!} = \frac{6!}{3!} = 6 \times 5 \times 4 = 120 \)
To find the total number of ways, we multiply the number of ways to seat the boys by the number of ways to seat the girls:
Total number of ways \( = 5! \times 6P_3 \)
\( = 120 \times 120 \)
\( = 14400 \)
So, there are 14,400 ways to seat 5 boys and 3 girls such that no two girls are together. This method ensures that at least one boy separates any two girls.
In simple words: First, arrange the boys. This creates empty spots between them. Then, place the girls in these spots so they are separated. Multiply the number of ways for boys and girls to get the final answer.

🎯 Exam Tip: For "no two together" problems, always arrange the items that *can* be together first, then place the other items in the gaps created by the first arrangement.

Question 3. How many 6-digit telephone numbers can be constructed with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each number starts with 35 and no digit appears more than once?
Answer: We need to form a 6-digit telephone number.
The problem states that the number must start with 35. This means the first two digits are fixed as 3 and 5.
So, the structure of the number is 3 5 _ _ _ _.
We have 10 available digits in total: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
Since the digits 3 and 5 are already used for the first two positions, and no digit can appear more than once (no repetition), we are left with \( 10 - 2 = 8 \) digits.
These 8 remaining digits are {0, 1, 2, 4, 6, 7, 8, 9}.
We need to fill the remaining 4 positions.
For the 3rd position, we have 8 choices (any of the remaining 8 digits).
For the 4th position, we have 7 choices (since one digit has been used for the 3rd position).
For the 5th position, we have 6 choices.
For the 6th position, we have 5 choices.
The total number of 6-digit telephone numbers is the product of the number of choices for each of the remaining 4 positions:
Number of 6-digit telephone numbers \( = 8 \times 7 \times 6 \times 5 \)
\( = 1680 \)
Thus, 1680 such unique telephone numbers can be created. This demonstrates how restrictions limit the available choices for each position.
In simple words: Two digits (3 and 5) are fixed at the start. Since digits cannot repeat, we have fewer choices for each of the next four spots. Multiply the number of choices for each spot to get the total number of phone numbers.

🎯 Exam Tip: When solving permutation problems with restrictions, first account for the fixed positions, then adjust the total number of available items and the number of positions to be filled for the remaining slots.

Question 4. Find the number of arrangements that can be made out of the letters of the word "ASSASSINATION".
Answer: To find the number of arrangements (permutations) of the letters in a word with repeated letters, we use the formula: \( \frac{n!}{n_1! n_2! ... n_k!} \), where n is the total number of letters and \( n_1, n_2, ..., n_k \) are the frequencies of each repeated letter.
First, let's count the total number of letters in "ASSASSINATION".
Total letters, \( n = 13 \).
Next, let's count the occurrences of each unique letter:
The letter A occurs 3 times.
The letter S occurs 4 times.
The letter I occurs 2 times.
The letter N occurs 2 times.
The letter T occurs 1 time.
The letter O occurs 1 time.
Now, we can apply the formula:
Number of arrangements \( = \frac{13!}{3! \times 4! \times 2! \times 2! \times 1! \times 1!} \)
This can be written as:
\[ \frac{13!}{3! 4! 2! 2! 1! 1!} = \frac{13!}{3! 4! 2! 2!} \]
Calculating the values:
\( 13! = 6,227,020,800 \)
\( 3! = 3 \times 2 \times 1 = 6 \)
\( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
\( 2! = 2 \times 1 = 2 \)
So, the denominator is \( 6 \times 24 \times 2 \times 2 = 576 \).
Number of arrangements \( = \frac{6,227,020,800}{576} = 10,810,800 \)
There are 10,810,800 distinct ways to arrange the letters of the word "ASSASSINATION". This formula accounts for the indistinguishable nature of identical letters.
In simple words: Count all the letters in the word. Then count how many times each letter appears. Use a special formula that divides the total factorial by the factorial of how many times each repeating letter shows up. This gives you all the unique ways to arrange the letters.

🎯 Exam Tip: Always double-check your letter counts for both the total number of letters and the frequency of each repeating letter to avoid calculation errors.

Question 5.
(a) In how many ways can 8 identical beads be strung on a necklace?
(b) In how many ways can 8 boys form a ring?
Answer:
(a) For identical items arranged in a circle, the number of distinct arrangements is given by \( \frac{(n-1)!}{2} \), because reversing the order does not create a new arrangement for identical items on a necklace (e.g., clockwise and anticlockwise arrangements are the same).
Here, \( n = 8 \) identical beads.
Number of ways \( = \frac{(8-1)!}{2} \)
\( = \frac{7!}{2} \)
\( = \frac{5040}{2} \)
\( = 2520 \)
There are 2520 ways to string 8 identical beads on a necklace. This rule specifically applies to symmetrical arrangements.
(b) For distinct items arranged in a circle (like boys forming a ring, as boys are distinct individuals), the number of arrangements is \( (n-1)! \). This is because one person's position is fixed to account for rotational symmetry, and the remaining \( n-1 \) people can be arranged in \( (n-1)! \) ways.
Here, \( n = 8 \) boys.
Number of ways \( = (8-1)! \)
\( = 7! \)
\( = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
\( = 5040 \)
There are 5040 ways for 8 boys to form a ring.
In simple words: (a) For beads that all look the same on a necklace, you find the ways to arrange them in a line, then halve it because flipping the necklace doesn't make a new pattern. (b) For different people in a circle, you fix one person's spot, and then arrange the rest in a line to find the total ways.

🎯 Exam Tip: Remember the distinction between identical items and distinct items when arranging them in a circle. Identical items (like beads) have an additional division by 2 to account for symmetry.

Question 6. Find the rank of the word 'CHAT' in the dictionary.
Answer: To find the rank of a word, we list its letters alphabetically and then count the permutations of words that come before it.
The letters of the word CHAT in alphabetical order are A, C, H, T.
The word we want to find the rank for is CHAT.

1. Words starting with A:
If 'A' is fixed as the first letter, the remaining letters (C, H, T) can be arranged in \( 3! \) ways.
\( 3! = 3 \times 2 \times 1 = 6 \) words. (ACHT, ACTH, AHCT, AHTC, ATCH, ATHC)

2. Words starting with C:
The second letter of CHAT is H. Since H comes after A in the alphabetical list of remaining letters for C (which are A, H, T), we consider words starting with CA.
Words starting with CA:
If 'CA' is fixed as the first two letters, the remaining letters (H, T) can be arranged in \( 2! \) ways.
\( 2! = 2 \times 1 = 2 \) words. (CAHT, CATH)

3. Words starting with CH:
The third letter of CHAT is A. Since A is the first letter alphabetically among the remaining for CH (which are A, T), we consider words starting with CHA.
Words starting with CHA:
If 'CHA' is fixed as the first three letters, the remaining letter (T) can be arranged in \( 1! \) way.
\( 1! = 1 \) word. (CHAT)
This is our target word.

Now, we sum up the counts of words before CHAT:
Rank of the word CHAT \( = \) (Words starting with A) \( + \) (Words starting with CA) \( + \) (Words starting with CHA and before CHAT)
\( = 3! + 2! + 1! \)
\( = 6 + 2 + 1 \)
\( = 9 \)
Therefore, the rank of the word 'CHAT' in a dictionary formed by its letters is 9. This method systematically accounts for all lexicographically smaller permutations.
In simple words: To find the rank, we count how many words come before "CHAT" if all the letters (A, C, H, T) were arranged in dictionary order. We count words starting with 'A', then 'C' and 'A', then 'C' and 'H' and 'A', until we reach "CHAT". Adding these counts gives the rank.

🎯 Exam Tip: When calculating dictionary rank, always list the available letters in alphabetical order first. Systematically count permutations for prefixes that come before the target word's prefix, letter by letter.