finding the size of a family with combinatorics

Here's a cool problem, not very difficult to solve, but there is a bit of number crunching.

Four children are picked at random (with no replacement) from a family which includes exactly two boys. The chance that neither boy is chosen is half the chance that both are chosen. How large is the family?

Any time you see 'choose' (order doesn't matter) and 'no replacement' together in the problem, you'll most likely use combinations to solve it. The easiest part of such a problem is to find the denominator of the fraction that gives the probability for an event expressed with combinatorics, which represents every event in the sample space. In this case, it's , where is the unknown number of people in the family. Then the number of ways to choose a group containing no boys is (i.e., choose four kids from the whole group minus the two boys), and the number of ways to choose a group containing both boys is . Actually, it's (i.e., choose two from the two boys and two from the remaining girls, who are out of all the kids), but is always one, so that can be left out. All told, the equation described in the original problem is . With some algebra, that reduces to , whose roots are two and seven. Since there's no way to take four from two (at least not in ), there are seven kids. Must be a Catholic family. And indeed .