A group contains 7 boys and some girls. The number of teams of 5 comprising 3 boys and 2 girls is 525. How many girls are in the group? 5 6 7 9 11

With 7 boys, you can choose 7C3 = 35 groups containing three boys. From n girls, you can choose nC2 = n(n-1)/2 teams containing two girls. We can thus choose (7C3)*(nC2) teams of five, containing three boys and two girls. We know that this number is equal to 525: 7C3*nC2 = 525 35*(n*(n-1))/2 = 525 n*(n-1)/2 = 15 n*(n-1) = 30 n = 6 (since n must be positive, n cannot be -5). So there are 6 girls.

Ian Stewart wrote:From n girls, you can choose nC2 = n(n-1)/2 teams containing two girls Thanks for the explanation Stewart! As usual, it's pragmatic and useful for the actual test! Just had a question about the above. How did you infer that equation above? Is that something that you just knew or do you have to know some specific combinatorics equation? I managed to get nC2 to... n! / ( 2!(n-2)! ) but couldn't simplify further.

jsl wrote:Ian Stewart wrote: Just had a question about the above. How did you infer that equation above? Is that something that you just knew or do you have to know some specific combinatorics equation? I managed to get nC2 to... n! / ( 2!(n-2)! ) but couldn't simplify further. No, I'm not using any special formula here - if you got to n! / ( 2!(n-2)! ), you were just about there. Whenever you divide one factorial by another, you'll always get a lot of cancellation. For example, (8!)/(6!) = 8*7, because 8! = 8*7*6! so (8!)/(6!) = (8*7*6!)/(6!) = 8*7 I was doing the same, but with n! and (n-2)!. Because n! = n*(n-1)*(n-2)! n!/(n-2)! = n*(n-1)*(n-2)!/(n-2)! = n*(n-1)

Thanks Ian! That's great - such a new and core concept for me that it's now in my notebook!

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