Question

Brent’s after-school game club has 12 members from which a six-member team is created. Miguel’s after-school sports club has 10 members from which a six-member team is created. Which student’s club has more possible combinations for his six-member team?

Answer 1

Answer: The group of 12 members.

Step-by-step explanation:

Since, the total possible combination of the 6 member team from the the group of 12 students = $12_C_6 = \frac{12!}{6!(12-6)!} = \frac{12!}{6!6!} = \frac{12\times 11\times 10\times 9\times 8\times 7\times 6!}{6!6!} = \frac{12\times 11\times 10\times 9\times 8\times 7}{720} = 924$

Also, the total possible combination of the 6 member team from the group of  10 students = $10_C_6 = \frac{10!}{6!(10-6)!} = \frac{10\times 9\times 8\times 7\times 6!}{6!4!} = \frac{10\times 9\times 8\times 7}{24} = 210$

Since, 924 > 210

Therefore, the 12 student's club has more possible combinations for his six-member team.

Answer 2

Answer:

Brent’s club has more possible team combinations because there are more members to choose from.

Step-by-step explanation: