Permutations and combinations — practice questions

Here, the order matters; being President is different from being Vice-President. We are arranging 3 people from a group of 8.

This is a permutation problem with $n=8$ and $r=3$.

Number of ways = $^8 P_3$

$^8 P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!}$

$= \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1}$

$= 8 \times 7 \times 6$

$= 336$

There are 336 different ways to fill the three positions.

(a) Order does not matter when choosing a team. This is a combination with $n=12$ and $r=5$.

Number of teams = $^{12} C_5 = \frac{12!}{5!(12-5)!} = \frac{12!}{5!7!}$

$= \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792$ teams.

(b) This is a multi-stage problem. We must choose 2 defenders AND 3 attackers (to make a team of 5).

Step 1: Choose 2 defenders from 5. Ways = $^5 C_2 = \frac{5!}{2!3!} = 10$.

Step 2: Choose 3 attackers from 7. Ways = $^7 C_3 = \frac{7!}{3!4!} = 35$.

Step 3: Since we need both events to happen, we multiply the results. Total teams = (Ways to choose defenders) $\times$ (Ways to choose attackers) $= 10 \times 35 = 350$ teams.