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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]

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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

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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.

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New post 10 Feb 2011, 07:15
I get it now... Thank you Bunuel!
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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]

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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]

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New post 07 May 2012, 14:11
Thanks a lot for the explanation Bunuel!!
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Re: A certain panel is to be composed of exactly three women and  [#permalink]

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New post 01 Jul 2013, 00:46
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Re: Selecting a panel  [#permalink]

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New post 03 Aug 2013, 06:07
Bunuel wrote:
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.


So in the same sense if we had (x-2)!*(x-1)!*3 in the denominator (x-2)!*(x-1)! would have been reduced?
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Re: A certain panel is to be composed of exactly three women and  [#permalink]

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New post 29 Oct 2013, 05:40
1
1
(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"
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Re: A certain panel is to be composed of exactly three women and  [#permalink]

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New post 09 May 2014, 12:10
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. :|
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Re: A certain panel is to be composed of exactly three women and  [#permalink]

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New post 10 May 2014, 05:56
yrt wrote:
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. :|


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.
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Re: Selecting a panel  [#permalink]

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New post 10 May 2014, 15:31
Bunuel wrote:
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.


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!
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Re: Selecting a panel  [#permalink]

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New post 11 May 2014, 06:18
russ9 wrote:
Bunuel wrote:
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.


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!


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.
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Re: Selecting a panel  [#permalink]

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New post 15 May 2014, 17:00
Bunuel wrote:
russ9 wrote:

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!


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.


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.
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Re: Selecting a panel  [#permalink]

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New post 16 May 2014, 01:11
2
russ9 wrote:

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.


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)*(x-1)*.....*1]
= (x+2)* (x+1)* [x!]
= (x+2)* (x+1)* x!

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Re: Selecting a panel  [#permalink]

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New post 19 May 2014, 20:36
gmatacequants wrote:
russ9 wrote:

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.


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)*(x-1)*.....*1]
= (x+2)* (x+1)* [x!]
= (x+2)* (x+1)* x!

Kudos is the best form of appreciation


Very clear - i was missing the highlighted part! Kudos to you. Thanks!
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Re: Selecting a panel  [#permalink]

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New post 03 Jun 2014, 09:20
Bunuel wrote:
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 \(x\): \(C^{3}_{x+2}=56\) --> \(\frac{(x+2)!}{(x+2-3)!*3!}=56\) --> \(\frac{(x+2)!}{(x-1)!*3!}=56\) --> \(\frac{x(x+1)(x+2)}{3!}=56\) --> \(x(x+1)(x+2)=6*7*8\) --> \(x=6\). << --- how to calculate roots fast for 3rd degree polynomial.

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

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

Answer: C.


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.
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Re: Selecting a panel  [#permalink]

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New post 03 Jun 2014, 09:33
PiyushK wrote:
Bunuel wrote:
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 \(x\): \(C^{3}_{x+2}=56\) --> \(\frac{(x+2)!}{(x+2-3)!*3!}=56\) --> \(\frac{(x+2)!}{(x-1)!*3!}=56\) --> \(\frac{x(x+1)(x+2)}{3!}=56\) --> \(x(x+1)(x+2)=6*7*8\) --> \(x=6\). << --- how to calculate roots fast for 3rd degree polynomial.

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

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

Answer: C.


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.


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.
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Re: Selecting a panel  [#permalink]

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New post 03 Jun 2014, 09:38
Bunuel wrote:
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.


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
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New post 03 Jun 2014, 09:41
PiyushK wrote:
Bunuel wrote:
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.


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


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.
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