GSEB Class 11 Maths Solutions Chapter 7 Permutations and Combinations Exercise 7.3

Gujarat Board Textbook Solutions Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3

 

Question 1. How many 3-digit numbers can be formed by using the digits 1 to 9, if no digit is repeated?
Answer: Three-digit numbers need to be created. This means three positions must be filled using the digits from 1 to 9. Since no digit is repeated, the choices decrease for each subsequent position. The first position has 9 options (digits 1-9), the second has 8 options (remaining 8 digits), and the third has 7 options (remaining 7 digits). This task can be achieved in \( 9 \times 8 \times 7 \) ways, which results in 504 options.
In simple words: To make 3-digit numbers without repeating any digits from 1 to 9, you have 9 choices for the first digit, 8 for the second, and 7 for the third. Multiply these together to get the total number of ways.

Exam Tip: Remember that "no digit is repeated" means you use permutations, and the number of available choices decreases for each position you fill.

 

Question 2. How many 4-digit numbers are there with no digit repeated?
Answer: To form 4-digit numbers using digits 0 to 9 without repetition, we must consider that the first digit cannot be zero. For the first position (thousands place), there are 9 possible digits (1 to 9). For the second position (hundreds place), there are also 9 options (0 and the remaining 8 digits). For the third place (tens place), there are 8 options left, and for the last place (units place), there are 7 options. Therefore, the total count of such numbers is calculated as: The number of permutations of 4 digits from 10 is \( ^{10}P_4 \). We subtract the numbers where 0 is in the first place, which are effectively 3-digit numbers formed from the remaining 9 digits, or \( ^{9}P_3 \). Thus, \( ^{10}P_4 - ^{9}P_3 = (10 \times 9 \times 8 \times 7) - (9 \times 8 \times 7) \) \( = 5040 - 504 = 4536 \) Alternatively, directly, the first digit has 9 choices (1-9), the second has 9 choices (0 and remaining 8), the third has 8 choices, and the fourth has 7 choices. So, \( 9 \times 9 \times 8 \times 7 = 4536 \).
In simple words: When making 4-digit numbers with no repeats, the first digit cannot be zero. So, you have 9 options for the first spot, 9 for the second (including zero now), 8 for the third, and 7 for the last. Multiply these choices to get the answer.

Exam Tip: Remember to account for the restriction of '0' not being allowed in the leading position for multi-digit number formation problems.

 

Question 3. How many 3-digit even numbers can be made, using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Answer: To make a 3-digit even number using the digits {1, 2, 3, 4, 6, 7} without repeating any, we first look at the last digit. An even number must end in 2, 4, or 6. So, there are 3 choices for the unit's place. After picking one digit for the unit's place, 5 digits are left from the original set of 6. We need to fill the remaining two places (hundreds and tens). The number of ways to pick and arrange 2 digits from the remaining 5 is \( ^{5}P_2 \). \( ^{5}P_2 = 5 \times 4 = 20 \) Since there are 3 options for the unit's place (2, 4, or 6), the total number of 3-digit even numbers possible is \( 3 \times 20 = 60 \).
In simple words: For 3-digit even numbers with no repeats, first choose the last digit (2, 4, or 6). Then, pick two more digits from the remaining ones for the first two spots. Multiply the choices for the last digit by the choices for the first two digits.

Exam Tip: When forming numbers with specific properties (like 'even'), always consider the restrictive conditions (e.g., unit's digit) first, then fill the remaining places.

 

Question 4. Find the 4-digit numbers that can be formed, using the digits 1, 2, 3, 4 and 5, if no digit is repeated. How many of these will be even?
Answer:
(i) Using the 5 digits {1, 2, 3, 4, 5}, we want to form 4-digit numbers without repeating any digits. The number of such arrangements is calculated using permutations: \( ^{5}P_4 \). This equals \( 5 \times 4 \times 3 \times 2 = 120 \). Therefore, 120 different 4-digit numbers can be formed.
(ii) For the 4-digit numbers to be even, the digit in the unit's place must be either 2 or 4. **Scenario 1: Unit's place is 2.** If 2 is used, 4 digits remain. The other three places can be filled in \( ^{4}P_3 \) ways. \( ^{4}P_3 = 4 \times 3 \times 2 = 24 \). **Scenario 2: Unit's place is 4.** If 4 is used, 4 digits remain. The other three places can be filled in \( ^{4}P_3 \) ways. \( ^{4}P_3 = 4 \times 3 \times 2 = 24 \). The total number of even 4-digit numbers is the sum of these scenarios: \( 24 + 24 = 48 \).
In simple words: First, count all possible 4-digit numbers from the given 5 digits with no repeats. Then, to find the even numbers, only allow 2 or 4 as the last digit. Calculate how many ways the other digits can be arranged for each case (ending in 2, then ending in 4) and add them up.

Exam Tip: Break down counting problems into independent cases when a specific condition (like 'even' or 'odd') applies to one position. Add the results of each case.

 

Question 5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman, assuming one person cannot hold more than one position?
Answer: To select a chairman from 8 persons, there are 8 possible ways. Once a chairman is chosen, 7 persons are left. From these remaining 7, a vice chairman can be picked in 7 ways. Since one person cannot hold both roles, we multiply the number of choices: \( 8 \times 7 = 56 \) ways. So, there are 56 ways to choose a chairman and a vice chairman.
In simple words: To pick a chairman and then a vice chairman from 8 people, you have 8 choices for the first role. After that, 7 people are left for the second role. Multiply these numbers together to find the total ways.

Exam Tip: Distinguish between permutations (where order matters, like selecting a chairman and vice-chairman) and combinations (where order does not matter).

 

Question 6. Find n, if \( ^{n-1}P_3 : ^nP_4 = 1 : 9 \).
Answer: We are given the ratio \( ^{n-1}P_3 : ^nP_4 = 1 : 9 \). This can be written as a fraction: \( \frac{^{n-1}P_3}{^nP_4} = \frac{1}{9} \). Using the permutation formula \( ^kP_r = \frac{k!}{(k-r)!} \), we can substitute the terms: \( \frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}} = \frac{1}{9} \) This simplifies to: \( \frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!} = \frac{1}{9} \) The \( (n-4)! \) terms cancel out: \( \frac{(n-1)!}{n!} = \frac{1}{9} \) We know that \( n! = n \times (n-1)! \). So, substitute this into the equation: \( \frac{(n-1)!}{n \times (n-1)!} = \frac{1}{9} \) The \( (n-1)! \) terms also cancel: \( \frac{1}{n} = \frac{1}{9} \) Therefore, \( n = 9 \).
In simple words: To solve this problem, write out the permutation formulas as a fraction. Then, simplify the factorials by canceling common terms until you get a simple equation to find 'n'.

Exam Tip: Simplify factorial expressions carefully. Remember that \( n! = n \times (n-1)! \) is key for canceling terms in permutation and combination problems.

 

Question 7. Find r, if
(i) \( ^5P_r = 2 \times ^6P_{r-1} \)
(ii) \( ^5P_r = ^6P_{r-1} \)
Answer:
(i) We need to find the value of r when \( ^{5}P_r = 2 \times ^{6}P_{r-1} \). Using the permutation formula \( ^{n}P_r = \frac{n!}{(n-r)!} \), we write: \( \frac{5!}{(5-r)!} = 2 \times \frac{6!}{(6-(r-1))!} \) Simplify the denominator on the right side: \( \frac{5!}{(5-r)!} = 2 \times \frac{6!}{(7-r)!} \) Expand \( 6! = 6 \times 5! \) and \( (7-r)! = (7-r)(6-r)(5-r)! \): \( \frac{5!}{(5-r)!} = 2 \times \frac{6 \times 5!}{(7-r)(6-r)(5-r)!} \) Cancel \( 5! \) and \( (5-r)! \) from both sides: \( 1 = \frac{12}{(7-r)(6-r)} \) Multiply both sides by \( (7-r)(6-r) \): \( (7-r)(6-r) = 12 \) Expand the left side: \( 42 - 7r - 6r + r^2 = 12 \) Rearrange into a quadratic equation: \( r^2 - 13r + 30 = 0 \) Factor the quadratic equation: \( (r-10)(r-3) = 0 \) This gives two possible values for r: \( r = 10 \) or \( r = 3 \). Since r cannot be greater than n (which is 5 in \( ^{5}P_r \)), the value \( r = 10 \) is not valid. Therefore, \( r = 3 \).
(ii) We need to find the value of r when \( ^{5}P_r = ^{6}P_{r-1} \). Using the permutation formula \( ^{n}P_r = \frac{n!}{(n-r)!} \), we write: \( \frac{5!}{(5-r)!} = \frac{6!}{(6-(r-1))!} \) Simplify the denominator on the right side: \( \frac{5!}{(5-r)!} = \frac{6!}{(7-r)!} \) Expand \( 6! = 6 \times 5! \) and \( (7-r)! = (7-r)(6-r)(5-r)! \): \( \frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r)(6-r)(5-r)!} \) Cancel \( 5! \) and \( (5-r)! \) from both sides: \( 1 = \frac{6}{(7-r)(6-r)} \) Multiply both sides by \( (7-r)(6-r) \): \( (7-r)(6-r) = 6 \) Expand the left side: \( 42 - 7r - 6r + r^2 = 6 \) Rearrange into a quadratic equation: \( r^2 - 13r + 36 = 0 \) Factor the quadratic equation: \( (r-9)(r-4) = 0 \) This gives two possible values for r: \( r = 9 \) or \( r = 4 \). Since r cannot be greater than n (which is 5 in \( ^{5}P_r \)), the value \( r = 9 \) is not valid. Therefore, \( r = 4 \).
In simple words: For both parts, use the permutation formula to set up an equation. Expand the factorials, cancel out common terms, and then solve the resulting quadratic equation for r. Always check that your solution for r is not bigger than 'n' in the permutation \( ^{n}P_r \).

Exam Tip: Always check the domain of r for permutations: \( 0 \le r \le n \). Discard any solutions for r that fall outside this range.

 

Question 8. How many words can be formed using all letters of the word EQUATION, using each letter exactly once?
Answer: The word EQUATION has 8 distinct letters. We need to find the number of ways to arrange all these letters, using each one exactly once. This is a permutation of 8 distinct items taken all at a time, which is represented by \( 8! \). \( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) \( 8! = 40320 \) Therefore, 40,320 different words can be formed.
In simple words: Since the word EQUATION has 8 unique letters and you use each one once to make new words, the number of ways is simply 8 multiplied by all the whole numbers smaller than it, down to 1 (which is 8 factorial).

Exam Tip: When arranging all distinct items, the number of permutations is simply \( n! \), where n is the total number of items.

 

Question 9. How many words, with or without meaning, can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
(i) 4 letters are used at a time?
(ii) all letters are used at a time?
(iii) all letters are used but first letter is a vowel?
Answer:
(i) The word MONDAY contains 6 unique letters. We need to find the number of ways to arrange 4 of these letters at a time. This is a permutation of 6 items taken 4 at a time, expressed as \( ^{6}P_4 \). \( ^{6}P_4 = 6 \times 5 \times 4 \times 3 = 360 \) So, 360 words can be formed using 4 letters at a time.
(ii) We need to arrange all 6 letters of the word MONDAY at a time. This is a permutation of 6 items taken all at a time, expressed as \( 6! \). \( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \) Thus, 720 words can be formed using all letters.
(iii) We need to form words using all letters of MONDAY, where the first letter must be a vowel. The vowels in MONDAY are A and O (2 vowels). **Case 1: The first letter is A.** After fixing 'A' in the first position, we have 5 remaining letters {M, O, N, D, Y} to arrange in the other 5 positions. This can be done in \( 5! \) ways. \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). **Case 2: The first letter is O.** Similarly, if 'O' is fixed in the first position, the remaining 5 letters can be arranged in \( 5! \) ways. \( 5! = 120 \). The total number of words starting with a vowel is the sum of these cases: \( 120 + 120 = 240 \).
In simple words: For part (i), pick 4 letters from 6 and arrange them. For part (ii), arrange all 6 letters. For part (iii), first choose a vowel for the start, then arrange the remaining letters in the remaining spots.

Exam Tip: When a condition specifies the starting (or ending) character, fix that position first, then calculate permutations for the remaining characters and positions.

 

Question 10. In how many of the distinct permutations of the letters in MISSISSIPPI, do the four 'I's not come together?
Answer: The word MISSISSIPPI has 11 letters. The frequency of letters is: I (4 times), S (4 times), P (2 times), M (1 time). **Step 1: Calculate total distinct permutations.** The total number of distinct permutations of these letters, without any restrictions, is given by the formula for permutations with repetitions: \( \frac{11!}{4!4!2!} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4! \times 4 \times 3 \times 2 \times 1 \times 2 \times 1} = 34650 \) **Step 2: Calculate permutations where all four 'I's come together.** To find arrangements where all four 'I's are together, we treat 'IIII' as a single block (one unit). Now we have 8 units to arrange: {M, S, S, S, S, P, P, (IIII)}. The repetitions are S (4 times) and P (2 times). The number of permutations for these 8 units is: \( \frac{8!}{4!2!} = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4! \times 2 \times 1} = 840 \) **Step 3: Calculate permutations where the four 'I's do not come together.** This is found by subtracting the 'I's-together' case from the total permutations: Number of permutations (I's not together) = Total permutations - Permutations (I's together) \( = 34650 - 840 = 33810 \) Therefore, there are 33,810 distinct permutations where the four 'I's do not appear consecutively.
In simple words: First, find all the possible ways to arrange the letters in MISSISSIPPI. Then, figure out how many of those arrangements have all four 'I's grouped together. Finally, subtract the "I's together" number from the total number of arrangements to find how many have the 'I's separated.

Exam Tip: For 'not together' problems, calculate the total arrangements and subtract the arrangements where the items *are* together. Treat the 'together' items as a single block.

 

Question 11. In how many ways, can the letters of the word PERMUTATIONS be arranged, if the
(i) Words start with P and end with S?
(ii) Vowels are together?
(iii) There are always 4 letters between P and S?
Answer: The word PERMUTATIONS has 12 letters, with the letter T appearing twice and all other letters being unique.
(i) We need to form words that start with P and end with S. **Step 1: Fix P and S.** Place 'P' at the first position and 'S' at the last position. This leaves 10 letters to arrange in the 10 middle positions. **Step 2: Arrange the remaining letters.** The remaining 10 letters are E, R, M, U, T, A, T, I, O, N. Among these, 'T' is repeated 2 times. The number of ways to arrange these 10 letters is \( \frac{10!}{2!} \). \( \frac{10!}{2!} = \frac{3,628,800}{2} = 1,814,400 \) Therefore, 1,814,400 words can be formed that start with P and end with S.
(ii) We need to arrange the letters of PERMUTATIONS such that all vowels stay together. **Step 1: Identify and group vowels.** The vowels are E, U, A, I, O (5 distinct vowels). Treat this group as a single unit. Within this unit, the vowels can be arranged in \( 5! \) ways. \( 5! = 120 \). **Step 2: Arrange the units.** The remaining letters are P, R, M, T, T, N, S (7 consonants). Now we have 7 consonants plus the vowel block, making a total of 8 units to arrange. Among these 8 units, the letter 'T' is repeated twice. The number of ways to arrange these 8 units is \( \frac{8!}{2!} \). \( \frac{8!}{2!} = \frac{40,320}{2} = 20,160 \) **Step 3: Combine the arrangements.** The total number of arrangements where vowels are together is the product of arranging the units and arranging the vowels within their block: \( 20,160 \times 120 = 2,419,200 \) So, 2,419,200 words can be formed where all vowels are together.
(iii) We need to arrange the letters of PERMUTATIONS such that there are always 4 letters between P and S. **Step 1: Determine positions for P and S.** There are 12 available positions. If P and S have 4 letters between them, this means P and S are separated by 5 positions (P _ _ _ _ S). Possible pairs of positions for (P, S): (1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11), (7, 12). There are 7 such pairs of positions. For each pair, P and S can be placed in two ways (P...S or S...P). So, the number of ways to place P and S is \( 7 \times 2 = 14 \).

**Step 2: Arrange the remaining letters.** After placing P and S, 10 positions remain, and 10 letters are left to fill them (E, R, M, U, T, A, T, I, O, N). Among these, 'T' is repeated twice. The number of ways to arrange these 10 letters in the remaining 10 places is \( \frac{10!}{2!} \). \( \frac{10!}{2!} = \frac{3,628,800}{2} = 1,814,400 \) **Step 3: Calculate the total arrangements.** Multiply the ways to place P and S by the ways to arrange the other letters: \( 14 \times 1,814,400 = 25,401,600 \) Therefore, 25,401,600 words can be formed with 4 letters always between P and S.
In simple words: For part (i), fix P at the start and S at the end, then arrange the 10 letters in between, remembering 'T' appears twice. For part (ii), group all vowels into one block, arrange that block with the consonants, and also arrange the vowels within their block. For part (iii), find all the possible places for P and S to have 4 letters between them (P....S or S....P), then arrange the remaining letters in the empty spots.

Exam Tip: For problems involving letters separated by a fixed number of other letters, first determine the possible starting positions for the pair, count the internal arrangements of that pair, and then arrange the remaining letters.