GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 May 2019, 20:07

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 30 Nov 2010
Posts: 209
Schools: UC Berkley, UCLA
A certain panel is to be composed of exactly three women and  [#permalink]

### Show Tags

09 Feb 2011, 14:05
5
21
00:00

Difficulty:

15% (low)

Question Stats:

74% (01:35) correct 26% (01:36) wrong based on 963 sessions

### HideShow timer Statistics

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

_________________
Thank you for your kudoses Everyone!!!

"It always seems impossible until its done."
-Nelson Mandela
Math Expert
Joined: 02 Sep 2009
Posts: 55150

### Show Tags

09 Feb 2011, 14:26
11
8
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.

Check this for more:
manhattan-q-ds-96244.html
defective-bulb-probability-99940.html
probability-of-2-different-representatives-89567.html
_________________
##### General Discussion
Manager
Joined: 30 Nov 2010
Posts: 209
Schools: UC Berkley, UCLA

### Show Tags

10 Feb 2011, 07:15
I get it now... Thank you Bunuel!
_________________
Thank you for your kudoses Everyone!!!

"It always seems impossible until its done."
-Nelson Mandela
Intern
Joined: 31 Jan 2012
Posts: 2
GMAT Date: 05-12-2012

### Show Tags

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?
Math Expert
Joined: 02 Sep 2009
Posts: 55150

### Show Tags

06 May 2012, 23:33
2
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?

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.
_________________
Intern
Joined: 31 Jan 2012
Posts: 2
GMAT Date: 05-12-2012

### Show Tags

07 May 2012, 14:11
Thanks a lot for the explanation Bunuel!!
Math Expert
Joined: 02 Sep 2009
Posts: 55150
Re: A certain panel is to be composed of exactly three women and  [#permalink]

### Show Tags

01 Jul 2013, 00:46
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

_________________
Manager
Joined: 26 Feb 2013
Posts: 150

### Show Tags

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?

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?
Intern
Joined: 04 Dec 2012
Posts: 11
GMAT Date: 12-26-2013
Re: A certain panel is to be composed of exactly three women and  [#permalink]

### Show Tags

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"
Intern
Joined: 03 Jan 2014
Posts: 1
Re: A certain panel is to be composed of exactly three women and  [#permalink]

### Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 55150
Re: A certain panel is to be composed of exactly three women and  [#permalink]

### Show Tags

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.
_________________
Manager
Joined: 15 Aug 2013
Posts: 243

### Show Tags

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?

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!
Math Expert
Joined: 02 Sep 2009
Posts: 55150

### Show Tags

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?

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.
_________________
Manager
Joined: 15 Aug 2013
Posts: 243

### Show Tags

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.
Intern
Joined: 13 May 2014
Posts: 33
Concentration: General Management, Strategy

### Show Tags

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!

Kudos is the best form of appreciation
Manager
Joined: 15 Aug 2013
Posts: 243

### Show Tags

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!
Director
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 750
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

### Show Tags

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.

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.
_________________
Piyush K
-----------------------
Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison
Don't forget to press--> Kudos
My Articles: 1. WOULD: when to use? | 2. All GMATPrep RCs (New)
Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".
Math Expert
Joined: 02 Sep 2009
Posts: 55150

### Show Tags

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.

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.
_________________
Director
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 750
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

### Show Tags

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
_________________
Piyush K
-----------------------
Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison
Don't forget to press--> Kudos
My Articles: 1. WOULD: when to use? | 2. All GMATPrep RCs (New)
Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".
Math Expert
Joined: 02 Sep 2009
Posts: 55150

### Show Tags

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.
_________________
Re: Selecting a panel   [#permalink] 03 Jun 2014, 09:41

Go to page    1   2    Next  [ 27 posts ]

Display posts from previous: Sort by