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A certain panel is to be composed of exactly three women and

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A certain panel is to be composed of exactly three women and [#permalink] New post 09 Feb 2011, 14:05
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A certain panel is to be composed of exactly three women and exactly two men, chosen from x women and y men. How many different panels can be formed with these constraints?

(1) If two more women were available for selection, exactly 56 different groups of three women could be selected.

(2) x = y + 1
[Reveal] Spoiler: OA

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Re: Selecting a panel [#permalink] New post 09 Feb 2011, 14:26
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mariyea wrote:
A certain panel is to be composed of exactly three women and exactly two men, chosen from x women and y men. How many different panels can be formed with these constraints?

(1) If two more women were available for selection, exactly 56 different groups of three women could be selected.

(2) x = y + 1


Question: C^{3}_{x}*C^{2}_{y}=? So we need the values of x and y.

(1) C^{3}_{x+2}=56 --> there is only one x to satisfy this (meaning that there is only one number out of which we can make 56 different selections of 3, so x+2 is some specific number which gives some specific x), but we know nothing about y. Not sufficient.

Just to illustrate how to find x: C^{3}_{x+2}=56 --> \frac{(x+2)!}{(x+2-3)!*3!}=56 --> \frac{(x+2)!}{(x-1)!*3!}=56 --> \frac{x(x+1)(x+2)}{3!}=56 --> x(x+1)(x+2)=6*7*8 --> x=6.

(2) x = y + 1. Clearly insufficient.

(1)+(2) From (1) we know x, so from (2) we can get y. Sufficient.

Answer: C.

Check this for more:
manhattan-q-ds-96244.html
defective-bulb-probability-99940.html
probability-of-2-different-representatives-89567.html
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Re: Selecting a panel [#permalink] New post 10 Feb 2011, 07:15
I get it now... Thank you Bunuel!
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Re: Selecting a panel [#permalink] New post 06 May 2012, 22:47
Hi Bunuel! In your explanation for statement 1, I understand that (x+2)! equals to (x+2)*(x+1)*x in the nominator but I don't understand how you got rid of (x-1)! in the denominator. Could you please explain?
Thanks for your help!!
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Re: Selecting a panel [#permalink] New post 06 May 2012, 23:33
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Hraisha wrote:
Hi Bunuel! In your explanation for statement 1, I understand that (x+2)! equals to (x+2)*(x+1)*x in the nominator but I don't understand how you got rid of (x-1)! in the denominator. Could you please explain?
Thanks for your help!!


Not so.

(x+2)!=1*2*3*...*(x-1)*x*(x+1)*(x+2)=(x-1)!*x*(x+1)*(x+2);

\frac{(x+2)!}{(x-1)!}=\frac{(x-1)!*x*(x+1)*(x+2)}{(x-1)!}=x*(x+1)*(x+2), as you can see (x-1)! is just being reduced.

Hope it's clear.
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Re: Selecting a panel [#permalink] New post 07 May 2012, 14:11
Thanks a lot for the explanation Bunuel!!
Re: Selecting a panel   [#permalink] 07 May 2012, 14:11
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