A committee of 3 ladies and 4 gents is to be formed out of 8 ladies and 7 gents. Mrs. X refuses to serve in a committee in which Mr. Y is a member. The number of such committees isa)1530b)1500c)1520d)1540Correct answer is option 'D'. Can you explain this answer?

Question

Answer

Problem Statement:

A committee of 3 ladies and 4 gents is to be formed out of 8 ladies and 7 gents. Mrs. X refuses to serve in a committee in which Mr. Y is a member. The number of such committees is:

Solution:

To form a committee of 3 ladies and 4 gents out of 8 ladies and 7 gents, we need to select 3 ladies out of 8 and 4 gents out of 7. The number of ways to do this is given by:

Number of ways to select 3 ladies out of 8 = 8C3 = (8*7*6)/(3*2*1) = 56

Number of ways to select 4 gents out of 7 = 7C4 = (7*6*5*4)/(4*3*2*1) = 35

The total number of ways to form a committee of 3 ladies and 4 gents is:

Total number of ways = 56 * 35 = 1960

Now, we need to subtract the number of committees in which Mrs. X and Mr. Y are both members. Let us denote the number of such committees by N.

To form a committee of 3 ladies and 4 gents out of 8 ladies and 7 gents, in which Mrs. X and Mr. Y are both members, we need to select 2 ladies out of 7 (excluding Mrs. X) and 3 gents out of 6 (excluding Mr. Y). The number of ways to do this is given by:

Number of ways to select 2 ladies out of 7 = 7C2 = (7*6)/(2*1) = 21

Number of ways to select 3 gents out of 6 = 6C3 = (6*5*4)/(3*2*1) = 20

The total number of ways to form a committee of 3 ladies and 4 gents, in which Mrs. X and Mr. Y are both members, is:

N = 21 * 20 = 420

Therefore, the number of committees in which Mrs. X and Mr. Y are not both members is:

Total number of ways - N = 1960 - 420 = 1540

Hence, the correct option is (d) 1540.

Answer

Answer

We are tasked with forming a committee of 3 ladies and 4 gents from 8 ladies and 7 gents, with the restriction that Mrs. X refuses to serve in a committee if Mr. Y is also a member.

Step 1: Total number of committees without restrictions
First, let's calculate the total number of committees that can be formed without any restrictions. The committee consists of 3 ladies and 4 gents, so we can choose 3 ladies from 8 and 4 gents from 7. This can be calculated as:

Number of ways to choose 3 ladies from 8 = ι₈³
Number of ways to choose 4 gents from 7 = ι₇⁴

Step 2: Apply the combination formula
Now, let's compute the values using the combination formula:
ι₈³ = (8 × 7 × 6) / (3 × 2 × 1) = 56
ι₇⁴ = (7 × 6 × 5 × 4) / (4 × 3 × 2 × 1) = 35

So, the total number of committees without restrictions is:
56 × 35 = 1960

Step 3: Committees where Mrs. X and Mr. Y are both members
Now, let's consider the cases where both Mrs. X and Mr. Y are on the committee. If Mrs. X is on the committee, we need to choose 2 more ladies from the remaining 7 ladies (since one lady, Mrs. X, is already chosen) and 3 more gents from the remaining 6 gents (since Mr. Y is already chosen). This can be calculated as:

Number of ways to choose 2 ladies from 7 = ι₇²
Number of ways to choose 3 gents from 6 = ι₆³

Step 4: Apply the combination formula
ι₇² = (7 × 6) / (2 × 1) = 21
ι₆³ = (6 × 5 × 4) / (3 × 2 × 1) = 20

So, the number of committees where both Mrs. X and Mr. Y are included is:
21 × 20 = 420

Step 5: Calculate the final number of valid committees
The number of valid committees (where Mrs. X and Mr. Y are not both members) is the total number of committees without restrictions minus the number of committees where both Mrs. X and Mr. Y are on the committee:

Valid committees = 1960 - 420 = 1540

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