Question

probability in real life
in this activity, you will apply your understanding of permutations and combinations to calculate probabilities. use the information in this scenario to answer the questions that follow.
coach bennets high school basketball team has 14 players, consisting of six juniors and eight seniors. coach bennet must select three players from the team to participate in a summer basketball clinic.
question 1
part a
how many different groups of three players are possible for coach bennet to select?
select the correct answer from each drop - down menu.
this is a combination/permutation because the order in which the players are selected is/isnt important.
there are different groups of three players possible for coach bennet to select.

Explanation:

Step1: Identify selection type

Since the order of selecting players does not matter (a group of players is the same regardless of selection order), this is a combination problem. So the first blank is "combination", and the second blank is "not".

Step2: Calculate total combinations

Use the combination formula $C(n,k)=\frac{n!}{k!(n-k)!}$, where $n=14$ (total players) and $k=3$ (players to select).

$$\begin{align*} C(14,3)&=\frac{14!}{3!(14-3)!}\\ &=\frac{14\times13\times12\times11!}{3\times2\times1\times11!}\\ &=\frac{14\times13\times12}{6}\\ &=364 \end{align*}$$

Answer:

This is a combination because the order in which the players are selected not important.
There are 364 different groups of three players possible for Coach Bennet to select.