It can help with a problem like this to keep an eye on the big picture, namely, that we are selecting 8 people from a group of 11 people, and without regard to order. So all the ways that this can happen, before we know anything about the make-up of the group and before we are given any restrictions, is 11C8, which is 165.
When we take a closer look, we see that this group consists of 4 males and 7 females. And we're told that the team that we're assembling needs at least 2 males and 2 females from these 11 people.
You may notice that there are lots of ways to distribute the people under this restriction, and this process could potentially involve lots of calculations. When you see this, it's often worthwhile to consider whether there is a quicker way.
Often the best way to count the ways something can happen is to count the ways it doesn't happen. In this case, how many ways does this restriction not get met? The answer, in general, is any situation where 0 or 1 males or 0 or 1 females are on the team.
If we look at how many males and females there are and compare them to the team numbers, we see that it is not possible to have 0 males or 0 females on the team. It is also impossible to have no more than 1 female on the team because there are only 4 males for a team of 8 people. The only way that the restriction doesn't get met is if all 7 females are on the team and only 1 male is. The number of ways this can happen is (7C7)(4C1), which equals 4.
So if we take all the possibilities and subtract the ways this doesn't happen, we get 165 - 4 = 161
Manel M.
Thank you so much! This made so much. I was initially trying to coming with all the possibilities of forming the group and adding the probabilities. For instance, the group could have 2 males and 6 females, 3 males and 5 females, 4 males and 4 females. I got a different result with this approach10/22/19