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PS - combination

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Manager
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PS - combination [#permalink] New post 16 Mar 2009, 05:44
00:00
A
B
C
D
E

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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
Expert Post
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Re: PS - combination [#permalink] New post 16 Mar 2009, 10:44
Expert's post
B
525 = C^7_3 * C^x_2=\frac{7*6*5}{3*2}*\frac{x*(x-1)}{2}

x(x-1)=30 ---> x=6
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Re: PS - combination [#permalink] New post 16 Mar 2009, 12:10
selvae wrote:
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


boys=> 7!/4!3!=35
525=35*g, where g is the number of teams comprising of girls only, so g=525/35=15 girls
We know that 15 = g!/(g-2)!2! , where g is the total number of girls available.

Solving backwards, and plugging the 1st available answer 5 quickly shows us that the number of girls should be a bit higher. Trying the number 6 works: 6!/4!2!=15

So, the answer is B.
Re: PS - combination   [#permalink] 16 Mar 2009, 12:10
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