CBSE Class 11 Mathematics Permutations And Combinations Worksheet Set 06

CBSE Class 11 Mathematics Worksheet - Permutations and Combinations (5). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. If \( ^nP_r = ^nP_{r+1} \) and \( ^nC_r = ^nC_{r+1} \), find the value of \( n \) & \( r \)?
Answer: We are given:
\( ^nP_r = ^nP_{r+1} \)
\( \Rightarrow \frac{n!}{(n-r)!} = \frac{n!}{(n-r-1)!} \)
\( \Rightarrow \frac{1}{(n-r)(n-r-1)!} = \frac{1}{(n-r-1)!} \)
\( \Rightarrow n - r = 1 \) .................... (1)

We also have:
\( ^nC_r = ^nC_{r+1} \)
\( \Rightarrow \frac{n!}{r!(n-r)!} = \frac{n!}{(r+1)!(n-r-1)!} \)
\( \Rightarrow \frac{1}{r!(n-r)(n-r-1)!} = \frac{1}{(r+1)r!(n-r-1)!} \)
\( \Rightarrow \frac{1}{n-r} = \frac{1}{r+1} \)
\( \Rightarrow n - r = r + 1 \) .................... (2)

From (1) and (2):
Substitute \( n - r = 1 \) into equation (2):
\( 1 = r + 1 \Rightarrow r = 0 \)
Substitute \( r = 0 \) into equation (1):
\( n - 0 = 1 \Rightarrow n = 1 \)
\( \therefore r = 0 \) & \( n = 1 \) ans.

Question. In an examination, a question paper consists of 12 questions divided into 2 parts, Part 1 & Part 2 containing 5 & 7 questions respectively. A student is required to attempt 8 questions in all, selecting at least 3 questions from each part. In how many ways can a student select questions?
Answer: We have Part 1 with 5 questions and Part 2 with 7 questions. The student needs to choose 8 questions in total, with at least 3 from each part. The possible cases are:

- Case 1: 3 questions from Part 1 & 5 questions from Part 2
Number of ways \( = {}^5C_3 \times {}^7C_5 = 10 \times 21 = 210 \)

- Case 2: 4 questions from Part 1 & 4 questions from Part 2
Number of ways \( = {}^5C_4 \times {}^7C_4 = 5 \times 35 = 175 \)

- Case 3: 5 questions from Part 1 & 3 questions from Part 2
Number of ways \( = {}^5C_5 \times {}^7C_3 = 1 \times 35 = 35 \)

\( \therefore \) Total number of ways to select questions \( = 210 + 175 + 35 = 420 \) ans.

Question. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Answer: In a deck of 52 cards, there are 4 aces and 48 non-ace cards.
To form a 5-card combination with exactly one ace, we must choose 1 ace from the 4 available aces, and 4 cards from the 48 non-ace cards:
Required number of combinations \( = {}^4C_1 \times {}^{48}C_4 \)
\( = 4 \times \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} \)
\( = 4 \times 194580 = 778320 \) ans.

Question. How many different words can be formed with the letters of the word MISSISSIPPI? In how many of these words do the 4 I's not come together?
Answer: In the word "MISSISSIPPI", total letters \( = 11 \) (M = 1, I = 4, S = 4, P = 2).

Total number of words:
\( \text{Total words} = \frac{11!}{4!4!2!} = 34650 \)

Number of words in which the 4 'I's are together:
Treating the 4 'I's as a single entity (IIII), we have 8 entities in total (M = 1, S = 4, P = 2, and 1 block of IIII):
\( \text{Arrangements with 'I's together} = \frac{8!}{4!2!} = 840 \)

Number of words in which the 4 'I's do not come together:
\( \text{Required words} = \text{Total words} - \text{Arrangements with 'I's together} \)
\( = \frac{11!}{4!4!2!} - \frac{8!}{4!2!} \)
\( = 34650 - 840 = 33810 \) ans.

Question. If \( ^nC_8 = ^nC_6 \), find \( ^nC_2 \)?
Answer: We know that if \( ^nC_x = ^nC_y \), then \( x + y = n \).
Given \( ^nC_8 = ^nC_6 \):
\( \Rightarrow n = 8 + 6 = 14 \)

Now, find \( ^nC_2 \):
\( ^nC_2 = {}^{14}C_2 = \frac{14 \times 13}{2 \times 1} = 91 \) ans.

Question. We wish to select 6 persons from 8, but if the person A is chosen, then B must be chosen. In how many ways can the selection be made?
Answer: We can divide the selection process into two mutually exclusive cases:

- Case 1: When person A is chosen:
Since A is chosen, B must also be chosen. Thus, 2 persons (A and B) are already selected. We now need to select the remaining 4 persons from the remaining 6 persons:
Number of ways \( = {}^6C_4 = {}^6C_2 = \frac{6 \times 5}{2} = 15 \).

- Case 2: When person A is not chosen:
Since A is excluded, we select 6 persons from the remaining 7 persons:
Number of ways \( = {}^7C_6 = 7 \).

\( \therefore \) Total number of ways to make the selection \( = 15 + 7 = 22 \) ans.

Question. How many words can be formed by taking 4 letters at a time from the letters of the word MURADABAD?
Answer: In the word "MURADABAD", total letters \( = 9 \) (A = 3, D = 2, M = 1, U = 1, R = 1, B = 1). There are 6 distinct letters available: {A, D, M, U, R, B}.
To form a 4-letter word, we consider the following cases:

- Case 1: All 4 letters are distinct:
Select 4 distinct letters from 6 in \( {}^6C_4 = 15 \) ways, and arrange them in \( 4! \) ways:
\( \text{Words} = 15 \times 24 = 360 \).

- Case 2: Two identical and two distinct (1 pair and 2 distinct letters):
The identical pair can be chosen from A or D (\( {}^2C_1 = 2 \) ways). The remaining 2 distinct letters are selected from the remaining 5 distinct letters (\( {}^5C_2 = 10 \) ways). Arrange them in \( \frac{4!}{2!} \) ways:
\( \text{Words} = 2 \times 10 \times 12 = 240 \).

- Case 3: Two pairs of identical letters (specifically, AA and DD):
Select both pairs in \( 1 \) way, and arrange them in \( \frac{4!}{2!2!} \) ways:
\( \text{Words} = 1 \times 6 = 6 \).

- Case 4: Three identical and one distinct (specifically, AAA and 1 distinct letter):
Select the triplet 'AAA' in 1 way, and choose 1 distinct letter from the remaining 5 distinct letters (\( {}^5C_1 = 5 \) ways). Arrange them in \( \frac{4!}{3!} \) ways:
\( \text{Words} = 1 \times 5 \times 4 = 20 \).

\( \therefore \) Total number of 4-letter words \( = 360 + 240 + 6 + 20 = 626 \) ans.

Question. There are 13 players out of which 4 are bowlers. In how many ways can a team of 11 be selected such that:
1. There is no restriction on selectors.
2. 1 particular player is never chosen.
3. 1 particular player is always chosen.
4. The team must include at least 3 bowlers.
Answer: We have a total of 13 players, which includes 4 bowlers and 9 non-bowlers.

1. There is no restriction on selectors:
Choose 11 players out of 13:
\( \text{Ways} = {}^{13}C_{11} = {}^{13}C_2 = \frac{13 \times 12}{2} = 78 \).

2. 1 particular player is never chosen:
Exclude the particular player. We select 11 players from the remaining 12 players:
\( \text{Ways} = {}^{12}C_{11} = 12 \).

3. 1 particular player is always chosen:
Fix the particular player in the team. We select the remaining 10 players from the remaining 12 players:
\( \text{Ways} = {}^{12}C_{10} = {}^{12}C_2 = 66 \).

4. The team must include at least 3 bowlers:
We can select either 3 or 4 bowlers:
- Case I (3 bowlers and 8 non-bowlers): \( {}^4C_3 \times {}^9C_8 = 4 \times 9 = 36 \)
- Case II (4 bowlers and 7 non-bowlers): \( {}^4C_4 \times {}^9C_7 = 1 \times 36 = 36 \)
\( \therefore \) Total ways \( = 36 + 36 = 72 \).

Question. A committee of 7 has to be formed from 9 boys & 4 girls. In how many ways can this be done, when the committee consists of:
1. Exactly 3 girls.
2. At least 3 girls.
3. At most 3 girls.
4. None of them is a girl.
5. Boys are in majority.
6. At least 3 from each gender.
7. 1 particular boy and 1 particular girl are never chosen.
Answer: We have 9 boys and 4 girls. We need to form a 7-member committee.

1. Exactly 3 girls:
Select 3 girls (out of 4) and 4 boys (out of 9):
\( \text{Ways} = {}^4C_3 \times {}^9C_4 = 4 \times 126 = 504 \).

2. At least 3 girls:
The committee can have 3 or 4 girls:
- Case I (3 girls, 4 boys): \( {}^4C_3 \times {}^9C_4 = 504 \)
- Case II (4 girls, 3 boys): \( {}^4C_4 \times {}^9C_3 = 1 \times 84 = 84 \)
\( \therefore \) Total ways \( = 504 + 84 = 588 \).

3. At most 3 girls:
The committee can have 0, 1, 2, or 3 girls:
- Case I (0 girls, 7 boys): \( {}^4C_0 \times {}^9C_7 = 1 \times 36 = 36 \)
- Case II (1 girl, 6 boys): \( {}^4C_1 \times {}^9C_6 = 4 \times 84 = 336 \)
- Case III (2 girls, 5 boys): \( {}^4C_2 \times {}^9C_5 = 6 \times 126 = 756 \)
- Case IV (3 girls, 4 boys): \( {}^4C_3 \times {}^9C_4 = 504 \)
\( \dots \) Total ways \( = 36 + 336 + 756 + 504 = 1632 \).

4. None of them is a girl:
Select 7 boys (out of 9):
\( \text{Ways} = {}^9C_7 = {}^9C_2 = 36 \).

5. Boys are in majority:
Boys are in majority if the committee has 4, 5, 6, or 7 boys, which corresponds to having at most 3 girls.
\( \therefore \) Total ways \( = 1632 \).

6. At least 3 from each gender:
The committee can have:
- 3 boys & 4 girls: \( {}^9C_3 \times {}^4C_4 = 84 \times 1 = 84 \)
- 4 boys & 3 girls: \( {}^9C_4 \times {}^4C_3 = 126 \times 4 = 504 \)
\( \therefore \) Total ways \( = 84 + 504 = 588 \).

7. 1 particular boy and 1 particular girl are never chosen:
Excluding the particular boy and girl, we have 8 boys and 3 girls (11 people total). We select 7 members from these 11:
\( \text{Ways} = {}^{11}C_7 = {}^{11}C_4 = 330 \).

Question. Find the no. of ways in which 5 boys and 5 girls be seated in a row so that:
1. No 2 girls may sit together.
2. All the girls are never together.
Answer:
1. No 2 girls may sit together:
First, arrange the 5 boys in \( 5! = 120 \) ways. This creates 6 gaps. We arrange the 5 girls in these 6 gaps in \( {}^6P_5 = 720 \) ways:
\( \text{Required ways} = 5! \times 6! = 120 \times 720 = 86400 \) ans.

2. All the girls are never together:
Total arrangements of 10 people in a row \( = 10! = 3628800 \).
Arrangements where all 5 girls sit together (treating the 5 girls as 1 block, giving 6 entities to arrange) \( = 6! \times 5! = 720 \times 120 = 86400 \).
\( \therefore \) Required ways \( = 10! - (5! \times 6!) = 3628800 - 86400 = 3542400 \) ans.

Question. A code word is to consist of 2 distinct English alphabets followed by 2 distinct numbers from 1 to 9. For example, CA23 is a code word. (i) how many such code words are there and (ii) how many of them end with an even integer?
Answer:
(i) Total code words:
The first 2 distinct letters out of 26 can be chosen and arranged in \( {}^{26}P_2 = 26 \times 25 = 650 \) ways. The 2 distinct digits out of 9 (from 1 to 9) can be chosen and arranged in \( {}^9P_2 = 9 \times 8 = 72 \) ways.
\( \therefore \) Total code words \( = 650 \times 72 = 46800 \) ans.

(ii) Ending with an even integer:
The last position must be filled by one of the 4 even numbers \( \{2, 4, 6, 8\} \) (4 ways). The third position can be filled by any of the remaining 8 distinct digits (8 ways). The first two letters can be filled in \( 650 \) ways.
\( \therefore \) Required code words \( = 650 \times 8 \times 4 = 20800 \) ans.

Question. A box contains 2 white balls, 3 black balls & 4 red balls. Find the number of ways of drawing 3 balls from the box if at least 1 black ball is to be included in the draw.
Answer: Total balls \( = 2 + 3 + 4 = 9 \) balls. Total non-black balls \( = 2 \text{ (white)} + 4 \text{ (red)} = 6 \) balls.
We can find the required ways by subtracting the ways of selecting only non-black balls from the total ways:
- Total ways to select 3 balls from 9 \( = {}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \).
- Ways to select 3 non-black balls \( = {}^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \).
\( \therefore \) Required ways \( = {}^9C_3 - {}^6C_3 = 84 - 20 = 64 \) ans.

Question. A committee of 6 is to be chosen from 10 men & 7 women so as to contain at least 3 men & at least 2 women. In how many different ways can this be done if two particular women refuse to serve on the same committee?
Answer: We have 10 men and 7 women. We need to choose a 6-member committee containing at least 3 men & at least 2 women.

1. Total committees containing at least 3 men & at least 2 women:
The possible combinations of (men, women) are:
- Case I (4 men and 2 women): \( {}^{10}C_4 \times {}^7C_2 = 210 \times 21 = 4410 \)
- Case II (3 men and 3 women): \( {}^{10}C_3 \times {}^7C_3 = 120 \times 35 = 4200 \)
Total ways \( = 4410 + 4200 = 8610 \).

2. Number of committees in which 2 particular women (say W1 and W2) serve together:
If W1 and W2 are fixed on the committee, we have:
- Case I (4 men and the 2 particular women): \( {}^{10}C_4 \times {}^2C_2 = 210 \times 1 = 210 \)
- Case II (3 men, the 2 particular women, and 1 other woman selected from the remaining 5 women): \( {}^{10}C_3 \times {}^2C_2 \times {}^5C_1 = 120 \times 5 = 600 \)
Total together \( = 210 + 600 = 810 \).

3. Required number of committees where they do not serve together:
\( \text{Required ways} = \text{Total} - \text{Together} = 8610 - 810 = 7800 \) ans.