Skill: Distinguish permutations from combinations

When counting, first ask:

Does the order of the objects matter?

• If different orders count as different outcomes, use a permutation:

$$nPr=\frac{n!}{(n-r)!}$$

• If different orders represent the same group, use a combination:

$$nCr=\frac{n!}{r!(n-r)!}$$

Here, $n$ is the total number of distinct objects, and $r$ is the number chosen.

Example

There are $6$ students: Alice, Ben, Carlos, Dana, Emma, and Farah.

Part 1: Choose 3 students for a committee

The committee $\{\text{Alice, Ben, Carlos}\}$ is the same group as $\{\text{Carlos, Alice, Ben}\}$. Therefore, order does not matter, so use a combination:

$$6C3=\frac{6!}{3!(6-3)!}$$ $$6C3=\frac{6!}{3!3!} =\frac{720}{6\cdot 6} =20$$

There are $\boxed{20}$ possible committees.

The factor $(3!)$ in the denominator removes the overcounting caused by arranging the same three students in different orders.

Part 2: Choose a president, vice president, and secretary

Now the roles are different. For example, Alice as president and Ben as vice president is different from Ben as president and Alice as vice president. Therefore, order matters, so use a permutation:

$$6P3=\frac{6!}{(6-3)!}$$ $$6P3=\frac{6!}{3!} =6\cdot5\cdot4 =120$$

There are $\boxed{120}$ possible assignments.

A useful check is that each group of $3$ students can be assigned to the three roles in $(3!)$ ways:

$$20\cdot 3!=20\cdot6=120$$

So, use a combination for an unordered group and a permutation when the positions or roles make the order important.