Probability

Question

There are 11 members on board of directors. If they must form a subcommittee of 3 members, how many subcommittees are possible?

I’m still unsure of this problem because a tutor suggested that there are 990 different subcommittees possible. And another tutor concluded that it was 165.

Answer 1

There are 3 spots to fill. 11 possibilities for the first spot, 10 left for the 2nd, and 9 left for the third. Multiplying these together gives 990 but this is not the right answer because in this problem order does not matter. We still have to divide by 3! to get rid of duplicate subcommittees. For example choosing Bob, Jill, and Mark results in the same subcommittee as choosing Mark, Bob and Jill. 990/3! = 165

One answer treats the problem like a permutation and the other treats it like a combination