We want to divide 10 dogs (including Fluffy and Nipper) into three groups, one with 3 dogs, one with 5 dogs, and one with 2 dogs. How many ways can we form the groups such that Fluffy is in the 2-dog group and Nipper is in the 5-dog group?
Alright, so here's the deal, I think I might be able to do this problem, but no guarantee that it is going to be right. I advise you to check my work afterward, to make sure that it is right:
So, we first know that there are 10 dogs, and we have to split them up into a 3 group, a 5 group, and a 2 group. However, one dog is already set in stone for the 2 dog group, as well as another dog set in stone for the 5 dog group.
We first pick the 3 dog group. We can do this by:
C(8,3) which means (choose 3 of 8). C(8,3) is 56.
Then, for the 5 dog group, we so far have 1 dog already established, then from the remaining 5 dogs left, we can pick 4 of them.
C(5,4) = 5 ways here.
The last group is already decided because we already picked the other two.
When we do 56 * 5 = 280, so I believe there are 280 ways to do this. (However, please REMEMBER, I might be incorrect [ and probably am] )