Case 1: What i did was i made "A" Included rather than B excluded which sounds like the same thing but when you do it
You have 8C5 comb? Because its guaranteed for 1 A so (10-1=9) which is 9 and B cant be in with A so 8 Choices. And when A fill in a spot There are 5 more spots left
8C5
That's correct. I'm not sure what your mistake is because if you did that then you should have arrived at the right answer.
The restriction is, essentially, that A and B cannot both be on the committee. You can have either one or the other but not both. A or B but not AB.
If you were to split the problem into cases in terms of "including A" then you would have:
Case 1: A is included. Then, B cannot be on the committee. You have five more spots to fill and two candidates have been eliminated so you get
.
Case 2: A is not included. There are no further restrictions here so you simply get
.
This question is also causing some probs for me
Im getting wayyyy offf for this
THe answer for this states 73080
We need exactly one independent, at least three liberal and at least one labor. We can break this up into three cases and calculate individually:
1 Ind 3 Lib 3 Lab
1 Ind 4 Lib 2 Lab
1 Ind 5 Lib 1 Lab
If you are faced with a problem where counting all the cases would not be viable, you might be able to take the complementary case instead. e.g. instead of "at least two" you could consider "one or zero" and subtract it from the number of permutations/combinations with no restrictions.