The problem states that a company manager forms a committee with 5 members. 6 people are chosen from a group of 12. Two people, A and B, are always included in the committee. The problem asks how many ways can the committee be formed?

1. Problem Description

The problem states that a company manager forms a committee with 5 members. 6 people are chosen from a group of

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2. Two people, A and B, are always included in the committee. The problem asks how many ways can the committee be formed?

2. Solution Steps

Since A and B are always included, we need to choose the remaining members from the remaining candidates.

Total number of members to choose is

5. A and B are already selected, so we need to select $5 - 2 = 3$ more members.

Initially, there were 12 candidates. Since A and B are included, they are no longer candidates to be chosen, so the number of candidates remaining is $12 - 2 = 10$.

We need to select 3 members from the remaining 10 candidates.

The number of ways to do this is given by the combination formula:

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

where n is the total number of items to choose from, and k is the number of items to choose.

In this case, $n = 10$ and $k = 3$.

$C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$

3. Final Answer

The committee can be formed in 120 ways.