There are 15 girls and 12 boys in a class. A committee of five is selected for a project.?

This a standard combinatorics problem.

Key things you need to know: C(n,k) = n!/[k!*(n-k)!], also denoted by an n over a k inside a big pair of parentheses, but I can't type that here, called "n choose k" or "the binomial coefficient n chose k" or "the number of combinations of n things k at a time" counts k element subsets of an n element set.

Here n! "n factorial" means n*(n-1)*...2*1 (unless n=0, in which case it's an empty product and thus equals 1).

If doing something can be thought of as doing part of the something then doing another part, and the number of ways to do each part doesn't depend on how the other part was done, then the number of ways to so the something is the number of ways to do the first part times the number of ways to so the second part [the great classic text Introductory Combinatoric by Brualdi calls this "the multiplicative principle"].

And if doing something can be thought of as doing one thing or another, the number of ways to do it is the number of ways to do the one thing plus the number of ways to do the other. ["the additive principle"]

So,three boys: to pick the committee you pick three out of 12 boys and pick two (5-3) out of 15 girls, so the answer is

C(12,3)*C(15,2) If you want a number in ordinary notation instead, you'll want to use a calculator -- if you're lucky you have one which has C(n,k), possibly denoted nCk or nCr, buried in some menu, otherwise you'll have to multiply and divide a lot.

At least one boy, means exactly one boy, or exactly two boys, or exactly three boys, or exactly four boys, or exactly five boys. I showed you how to find a the number of ways to pick a committee with exactly three boys, so use the same steps for 1, 2, 3, 4, and 5 instead and add up the results to get the answer to b.