Question

help.

Answer 1

with?

Answer 2

number 5.

Answer 3

@kirbykirby

Answer 4

\[{9 \choose 7}=36\]

Answer 5

oh i thought you multiplied 9 x 7.

Answer 6

Not quite. What you are doing is selecting a subset of 7 members from a total of 9.

And use a combination rather than a permutation since the order doesn't matter (e.g. it doesn't matter if (John, Alice) or (Alice, John) are in the committee)

Answer 7

The fact that you are selecting a subset from the larger set is important. It is in fact why you use a combination.

Answer 8

Maybe to convince more.. Think of this.... suppose they were selecting 8 out of the 9 members. How many committees can you form? It would not be 9*8, but again \(\large {9 \choose 8}=9\).

Say you have members
A B C D E F G H I

You can form groups:
A, B, ..., H
B, C, ...., I
C, D, ...., I, A
D, E, ...., I, A, B

to get 8 different groups

Answer 9

Sometimes counting arguments can be tricky. If you can reduce the problem to smaller numbers and figure out the "logic" behind it, and then generalize it to larger numbers, it can help if you are unsure which formula to use.

Answer 10

I'm trying to understand.