Question
Six friends are going to play a ball game. Each team has 3 players. How many different team combinations are possible?
(1 Point)
6
18
20
120
(1 Point)
6
18
20
120
Answers
GPT-4o mini
To determine how many different team combinations are possible when forming two teams of 3 players each from 6 friends, we can use the combination formula, which is given by:
\[
\binom{n}{r} = \frac{n!}{r!(n - r)!}
\]
where \( n \) is the total number of items (in this case, friends), \( r \) is the number of items to choose (team members), and \( ! \) denotes factorial.
In this case, we want to choose 3 players from 6. Thus, we have:
\[
\binom{6}{3} = \frac{6!}{3!(6 - 3)!} = \frac{6!}{3! \cdot 3!}
\]
Calculating this:
\[
6! = 720, \quad 3! = 6
\]
So,
\[
\binom{6}{3} = \frac{720}{6 \cdot 6} = \frac{720}{36} = 20
\]
However, please note that selecting Team A (e.g., players 1, 2, and 3) and Team B (players 4, 5, and 6) is the same as selecting Team B and Team A. Therefore, we need to divide by 2 to account for this overcounting:
\[
\text{Different team combinations} = \frac{\binom{6}{3}}{2} = \frac{20}{2} = 10
\]
But since the question is asking for how many different ways we can form a team regardless of the other team, we actually have calculated the combination for one team.
So the answer to how many different teams of 3 can be formed from 6 friends is **20**.
Thus, the correct answer is \(20\).
\[
\binom{n}{r} = \frac{n!}{r!(n - r)!}
\]
where \( n \) is the total number of items (in this case, friends), \( r \) is the number of items to choose (team members), and \( ! \) denotes factorial.
In this case, we want to choose 3 players from 6. Thus, we have:
\[
\binom{6}{3} = \frac{6!}{3!(6 - 3)!} = \frac{6!}{3! \cdot 3!}
\]
Calculating this:
\[
6! = 720, \quad 3! = 6
\]
So,
\[
\binom{6}{3} = \frac{720}{6 \cdot 6} = \frac{720}{36} = 20
\]
However, please note that selecting Team A (e.g., players 1, 2, and 3) and Team B (players 4, 5, and 6) is the same as selecting Team B and Team A. Therefore, we need to divide by 2 to account for this overcounting:
\[
\text{Different team combinations} = \frac{\binom{6}{3}}{2} = \frac{20}{2} = 10
\]
But since the question is asking for how many different ways we can form a team regardless of the other team, we actually have calculated the combination for one team.
So the answer to how many different teams of 3 can be formed from 6 friends is **20**.
Thus, the correct answer is \(20\).