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

Question

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

Bunuel · Accepted Answer

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

Bunuel · Answer

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.

mariyea · Answer

I get it now... Thank you Bunuel!

Hraisha · Answer

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!!

Hraisha · Answer

Thanks a lot for the explanation Bunuel!!

Bunuel · Answer

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Skag55 · Answer

So in the same sense if we had (x-2)!*(x-1)!*3 in the denominator (x-2)!*(x-1)! would have been reduced?

vinayak4u · Answer

(1) NC3=56
Only way this is possible is if N=8 i.e, 8*7*6/3! = 56

As N=X+2 => X=6 ; No info on Y, thus insufficient

(2) No values provided. Thus insufficient.

(1)+(2) => X=6;Y=5

Therefore "C"

yrt · Answer

I do not feel right about Bunuel's explanation for (1) option. 56 different groups of 3 women would include 2 men also right. so (x+2 C 3)*(y C 2) = 56 but not (x+2 C 3 = 56). Am I right.

Bunuel · Answer

No, you are not right.

(1) says: IF two more women were available for selection, exactly  56 different groups of three women could be selected . This hypothetical situation talks only about selection of three women IF two more women were available.

russ9 · Answer

Hi Bunuel,

As the other poster, I'm having a hard time following your simplification method.

Isn't (x+2)! = (x+2)(x+1)(x) ? Where are the negatives and single integers coming from? Thanks!

Bunuel · Answer

Factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n. For instance 5!=1*2*3*4*5. Hence the factorial of (x+2) is (x+2)!=1*2*3*...*(x-1)*x*(x+1)*(x+2)=(x-1)!*x*(x+1)*(x+2) . Notice that 1*2*3*...*(x-1)*x*(x+1) is factorial of (x+1), so (x+1)!. Where do you see negative numbers there? For more check here: math-number-theory-88376.html Hope it helps.

russ9 · Answer

Hi Bunuel,

I read the math theory page and i'm not seeing a section for factorials of an equation such as (x+2).

I understand the process that n! = N*(n-1)*(n-2)...

By that token - wouldn't (x+2)! be x*(x+1)*(x+2)? Why are you going on the other end of the number line essentially and more importantly, how are the 1*2*3 popping up? Also, what I meant by negative numbers is the expression (x  -1  )!

Thanks.

gmatacequants · Answer

Russ9 :
Factorial n denoted by n! is equal to multiplication of all integers starting from 1 till n.
So,notation n! is a shortcut to the expression : n*(n-1)*(n-2)*(n-3)*....3*2*1.
Now,
 (x+2)!  =  (x+2)*(x+1)*(x)*(x-1)*......*1 
 =  (x+2)* (x+1) *     
 =  (x+2)* (x+1)*   
 =  (x+2)* (x+1)* x!

Kudos is the best form of appreciation

russ9 · Answer

Very clear - i was missing the highlighted part! Kudos to you. Thanks!

PiyushK · Answer

Hi Bunuel,

How you calculated the value of X for 3rd degree polynomial ? Is there any test for third or above degree polynomial to guess signs of the roots without solving equation ? (Whether polynomial has 0 one of the root or n number of negative or positive roots )

Thanks.

Bunuel · Answer

I did not solve. 336 is the product of three consecutive positive integers, then I broke 336 into 6*7*8 making x to be 6. As I wrote in my solution you don't need to solve, you just need to know that you can solve and get x.

PiyushK · Answer

What if in case we have equation for which there are 2 or more real roots ? Is there any way to just look at the signs of coefficients and predict signs of the roots?
I tried to google out my query but failed to get any answer.

Thanks

Bunuel · Answer

You don't need to know how to solve 3rd or higher degree equations for the GMAT and you don't need this for this particular question either.

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

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