I have 10 people with the letter 'A' written on the front of them and 5 people with the letter 'B' written on the front of them. I select a group of 4 out of the total 15 people.
How many different groups can I make with the 15 people?
How many groups contain at least two people with 'A' on them?
So far I've got the total amount of combinations of 4 people, which is C(15, 4) = 1365 different possibilities, so that answers question 1.
How do I figure out the second question? From what I've read, the easiest way is to get the total amount of combinations and then subtract the complement of the second part (eg. how many groups of people DON'T contain two A's), but I'm stuck on how to find that equation.
I've tried subtracting C(10, 2) from the total because we assume the first two people have 'A' on them, so now we need to choose the remaining 2 people from 10 left over.. But I don't think that's right because the question says AT LEAST and not exactly.. Can anyone point me in the right direction please?
Best Answer
The complement should be either no one has 'A' on them or exactly one has 'A' on them.
Number of group where no one has 'A': $$\binom{5}{4}$$
Number of group where exactly one has 'A':
$$\binom{5}{3}\cdot \binom{10}{1}$$