# Homework Help: Simple Combination Help. Confused!

1. Feb 27, 2010

### Happyzor

1. The problem statement, all variables and given/known data
You have 7 birds lined up to feed. Only 3 birds can feed at a time. Two of the 7 birds do not like to feed with each other. How many combinaions can be formed?

2. Relevant equations
C(n,k)=n!/k!(n-k)!

3. The attempt at a solution
Attempted solution
C(7,3)-C(7,2)=21.

Can someone explain this to me? Thanks. My thinking was 7 choose 3 minus 7 choose the two birds that don't like to mix. Subtract and you'll get the birds that mix. Obviously its wrong -_-. Thanks for your help in advance.

2. Feb 27, 2010

### VeeEight

C(7,3) is clear.
Then you must subtract the number of times when the two birds are with each other. You must fix these two birds - that is where C(2,2) comes from (ie, these is only 1 way to select these two birds). The C(5,1) term comes from the fact that after you have selected these two birds, you must pick 1 last one so you have 3 on the feeding line.

Also, note that if 3 is the maximum number of birds allowed on the feeding line, then you can still consider situations when there are just 1 and 2 birds (respectively) on the line.

3. Feb 27, 2010

### Happyzor

Ah ok, thanks. I was thinking order mattered and all, but now its clear order doesn't matter. There is only one way to choose the two birds. Then the 3rd combination is the combinations with the other 5 birds. Thanks a bunch for clearing it up.