Question

A team leader needs to form a committee of 5 members from a group of 9 employees. However, two particular employees, Reena and Tina, conflict with each other and cannot work together in the committee. In how many ways can the team leader form a committee of 5 members?

• A. 35
• B. 70
• C. 56
• D. 91

Answer 1

Answer: (D)

91

Answer 2

Let the total number of employees be $N = 9$. The size of the committee to be formed is $K = 5$. Let the two particular employees who cannot work together be Reena (R) and Tina (T).

We need to find the number of ways to form a committee of 5 members from 9 such that Reena and Tina are not both in the committee.

We can solve this by finding the total number of ways to form the committee without any restrictions and subtracting the number of ways where both Reena and Tina are in the committee.

1. Total number of ways to form a committee of 5 from 9 employees without any restrictions: This is given by the combination formula $C(n, r) = \frac{n!}{r!(n-r)!}$, where $n$ is the total number of items, and $r$ is the number of items to choose. Total ways = $C(9, 5) = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!}$ $C(9, 5) = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5! \times 4 \times 3 \times 2 \times 1} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 9 \times 2 \times 7 = 126$.

2. Number of ways where both Reena and Tina are in the committee: If both Reena and Tina are already selected for the committee, we need to select the remaining $5 - 2 = 3$ members from the remaining $9 - 2 = 7$ employees. Number of ways with both R and T = $C(7, 3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!}$ $C(7, 3) = \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = 35$.

3. Number of ways where Reena and Tina are not both in the committee: This is the total number of ways minus the number of ways where both Reena and Tina are in the committee. Number of valid ways = Total ways - Ways with both R and T Number of valid ways = $C(9, 5) - C(7, 3) = 126 - 35 = 91$.

Therefore, the number of ways the team leader can form a committee of 5 members such that Reena and Tina cannot work together is 91.