A company manager wants to form a committee from 12 staff members. The committee must have 4 members. Among the 12 staff, there are 2 people, A and B, who must either both be in the committee or both not be in the committee. How many ways can such a committee be formed?

1. Problem Description

A company manager wants to form a committee from 12 staff members. The committee must have 4 members. Among the 12 staff, there are 2 people, A and B, who must either both be in the committee or both not be in the committee. How many ways can such a committee be formed?

2. Solution Steps

We will consider two cases:

Case 1: Both A and B are in the committee.

Since A and B are already in the committee, we need to choose 2 more members from the remaining $12 - 2 = 10$ staff members. The number of ways to do this is $\binom{10}{2}$.

Case 2: Neither A nor B is in the committee.

Since A and B are not in the committee, we need to choose 4 members from the remaining $12 - 2 = 10$ staff members. The number of ways to do this is $\binom{10}{4}$.

The total number of ways to form the committee is the sum of the number of ways in each case.

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

$\binom{10}{2} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45$

$\binom{10}{4} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210$

Total number of ways = $\binom{10}{2} + \binom{10}{4} = 45 + 210 = 255$

3. Final Answer

255