Number of ways to form two different committees

Question

There are five employees willing to serve on one of two different committees. If each employee can only serve on one committee, how many possible ways are there for the openings on the committees to be filled?

The answer is 20. And the explanation:

There are 5 ways for the opening on the first committee to be filled because all 5 people are willing to serve on either committee. With that opening filled, there are 4 ways for the second opening to be filled. There are 5•4=20 total ways to fill the opening. So the answer is 20.

Would someone please explain how the answer is $5\cdot4$

I don't understand it conceptually.

Answer 1

The answer should be $2^5$. Each person has a choice to go to committee $A$ or $B$. So $5$ of them would have a total of $2^5$ combinations.

Answer 2

I have three ways to explain the result.

a)

You have two committees. In each committee there has to be one of the 5 employees. The employees can be numbered 1,2,3,4,5. Now you can fill each of the two committee with an employee. Here is the first number the employee, which is in committee 1. And is the second number the employee, which is in committee 2.

$12 \quad $ $21\quad$ $13\quad$ $31\quad$ $14\quad$ $41\quad$ $15\quad$ $51\quad$ $23\quad$ $32\quad$ $24\quad$ $42\quad$ $25\quad$ $52\quad$ $34\quad$ $43\quad$ $35\quad$ $53\quad$ $45\quad$ $54\quad$

All together there are 20 ways.

b)

The explanation, which you quoted is just fine. First you choose one of the five employees for the first committee. There are 5 ways. 4 employees remaining. Thus you have 4 ways to fill the second committee with an employee.

c)

Or you can use the formula for permutation without repition. In permutations orders matters.

${n \choose k} \cdot k!={5 \choose 2} \cdot 2!=\frac{5 \cdot 4}{1 \cdot 2}\cdot 2=20$