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A basketball team has 10 players. Before each game, the coach picks 2 of those players to carry the team's water jug.
How many different groups of 2 players can the coach pick?

A basketball team has $10$ players. Before each game, the coach picks $2$ of those players to carry the team's water jug. $\newline$ How many different groups of $2$ players can the coach pick?

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Counting principle

Full solution

Q. A basketball team has

10

players. Before each game, the coach picks

2

of those players to carry the team's water jug.

\newline

How many different groups of

2

players can the coach pick?

Define Combination Formula: To determine the number of different groups of $2$ players that can be picked from $10$ players, we need to use the combination formula, which is defined as $C(n, k) = \frac{n!}{k!(n - k)!}$ , where $n$ is the total number of items, $k$ is the number of items to choose, and “ $!$ ” denotes factorial. $\newline$ In this case, $n = 10$ (total players) and $k = 2$ (players to be picked).
Calculate Factorial of $n$ : First, we calculate the factorial of $n$ , which is $10!$ ( $10$ factorial). $\newline$ $10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
Calculate Factorial of $k$ : Next, we calculate the factorial of $k$ , which is $2!$ ( $2$ factorial). $\newline$ $2! = 2 \times 1$
Calculate Factorial of $(n - k)$ : We also need to calculate the factorial of $(n - k)$ , which is $(10 - 2)!$ or $8!$ ( $8$ factorial). $\newline$ $8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
Apply Combination Formula: Now we can plug these values into the combination formula: $\newline$ $C(10, 2) = \frac{10!}{(2!(10 - 2)!)}$ $\newline$ $C(10, 2) = \frac{10!}{(2! \times 8!)}$
Simplify Factorials: We simplify the factorials by canceling out the common terms in $10!$ and $8!$ : $\newline$ $C(10, 2) = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!}$ $\newline$ $C(10, 2) = \frac{10 \times 9}{2 \times 1}$
Find Number of Combinations: Perform the division and multiplication to find the number of combinations: $\newline$ $C(10, 2) = \frac{10 \times 9}{2}$ $\newline$ $C(10, 2) = \frac{90}{2}$ $\newline$ $C(10, 2) = 45$

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