A manager of a company wants to form a committee with 5 members. There are 12 candidates. Two candidates A and B can be members of the committee. How many ways can the committee be formed? (I am guessing that A and B are always included, because I see "A & B អាចជាសមាជិកបាន" which means "A & B can be members".)

1. Problem Description

A manager of a company wants to form a committee with 5 members. There are 12 candidates. Two candidates A and B can be members of the committee. How many ways can the committee be formed? (I am guessing that A and B are always included, because I see "A & B អាចជាសមាជិកបាន" which means "A & B can be members".)

2. Solution Steps

The question is unclear. Let's consider two cases:

Case 1: A and B must be included in the committee. Since A and B are already in the committee, we need to choose 3 more members from the remaining 10 candidates.

The number of ways to choose 3 members from 10 candidates is given by the combination formula:

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

where $n$ is the total number of items, and $k$ is the number of items to choose. In this case, $n = 10$ and $k = 3$.

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

Case 2: Either A or B can be included or neither of them can be included.

This is not the interpretation of the problem as written in the picture.

Case 3:

The committee must have A or B or both.

Number of committees with only A: we have to choose 4 from

1

0. $C(10,4) = \frac{10*9*8*7}{4*3*2*1} = 210$

Number of committees with only B: we have to choose 4 from

1

0. $C(10,4) = \frac{10*9*8*7}{4*3*2*1} = 210$

Number of committees with A and B: we have to choose 3 from

1

0. $C(10,3) = \frac{10*9*8}{3*2*1} = 120$

Since we double counted number of committees with A and B if we directly add the result from committees with only A, only B, and A and B. We have to subtract $C(10,3)$ once, but we didn't count it initially, so we just have to add it.

Total = $C(10,4) + C(10,4) + C(10,3) = 210 + 210 + 120 = 540$.

Final case is where we can have any 5 of the 12 selected.

Then we have $C(12,5) = \frac{12!}{5!7!} = \frac{12*11*10*9*8}{5*4*3*2*1} = 12*11*3*2 = 792$

3. Final Answer

If A and B must be included in the committee, then there are 120 ways to form the committee.

If the problem allows any members, then there are $C(12,5) = 792$ number of ways.

Assuming A and B must be included, the final answer is 120.