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Question
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. part a question 1 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 because the order in which the players are selected there are different groups of three players possible for coach bennet to select.
Step1: Identify combination formula
The formula for combinations is $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n$ is the total number of items and $r$ is the number of items to be chosen. Here, $n = 14$ (total players) and $r=3$ (players to be selected).
Step2: Calculate factorial values
$n!=14! = 14\times13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1$, $r!=3!=3\times2\times1$, and $(n - r)!=(14 - 3)!=11!=11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1$. Then $C(14,3)=\frac{14!}{3!(14 - 3)!}=\frac{14\times13\times12\times11!}{3\times2\times1\times11!}$.
Step3: Simplify the expression
Cancel out the $11!$ terms. We get $C(14,3)=\frac{14\times13\times12}{3\times2\times1}=\frac{2184}{6}=364$.
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364