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.
A group contains 7 boys and some girls
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jsl
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Thanks for the explanation Stewart! As usual, it's pragmatic and useful for the actual test!Ian Stewart wrote:From n girls, you can choose nC2 = n(n-1)/2 teams containing two girls
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.
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Ian Stewart
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jsl wrote: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, becauseIan 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.
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)
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
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