To determine which student’s club has more possible combinations for their six-member teams, we can calculate the number of combinations for each club using the combinations formula:
\[ C(n, r) = \frac{n!}{r!(n - r)!} \]
where \( C(n, r) \) is the number of combinations, \( n \) is the total number of members in the club, and \( r \) is the number of members to be chosen for the team.
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Brent’s club:
- Total members (\( n \)) = 12
- Members chosen for the team (\( r \)) = 6
\[ C(12, 6) = \frac{12!}{6! \cdot (12 - 6)!} = \frac{12!}{6! \cdot 6!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924 \]
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Miguel’s club:
- Total members (\( n \)) = 10
- Members chosen for the team (\( r \)) = 6
\[ C(10, 6) = \frac{10!}{6! \cdot (10 - 6)!} = \frac{10!}{6! \cdot 4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]
Now, let's compare the results:
- Brent’s club: 924 combinations
- Miguel’s club: 210 combinations
Based on this analysis, Brent’s club has more possible combinations for his six-member team.
The correct conclusion is: "Brent’s club has more possible team combinations because there are more members to choose from."
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