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A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
29 Jul 2010, 19:36

3

This post received KUDOS

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

85% (hard)

Question Stats:

55% (03:05) correct
45% (02:34) wrong based on 104 sessions

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

Re: Another Combination Problem [#permalink]
29 Jul 2010, 19:41

7

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

jakolik wrote:

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\)

Ways to chose 6 members committee with two particular men serve together: \(C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65\)

Re: Another Combination Problem [#permalink]
15 Sep 2012, 07:28

4

This post received KUDOS

stne wrote:

Yups Like others I too have the same query

Anybody explain the logic selecting two men who work together with each other , please?

We only have the symbols here but nothing explaining the same ?

Quote:

Ways to chose 6 members committee with two particular men serve together: \(C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65\)

Denote by A and B the two particular men. We can have 2 men and 4 women chosen, when the two men are the particular two: choose the two men \(C^2_2\), then choose 4 women out of the 5, which is \(C^4_5\). So, A,B and four women chosen. We can also have the two men chosen, again \(C^2_2\), choose another man from the remaining 6 men, this is \(C^1_6\), then choose 3 women out of the 5, which is \(C^3_5.\) Here, A, B, another men, and 3 women are chosen. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
24 Sep 2012, 03:10

2

This post received KUDOS

Find out how many different combinations can be made between the 8 men and 5 women ...

We will have two scenarios , First where we have 2 men and 4 women , and second where we have 3 men and 4 women ... Finding the combinations for each and then adding them up gives us 700 ...

Logically the number of combinations where any 2 men do not serve together would have to be lower than 700 , therefore the only option is E .... _________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
15 Sep 2012, 07:48

1

This post received KUDOS

jakolik wrote:

Thank you Bunuel.

Can you please explain why ways to chose 6 members committee without restriction cannot be: C^2_8*C^3_5*C^1_8 We are to choose 2 men out of 8, 3 women out of 5 and 1 (woman or man) out of 8 (13-5)

Bunuel wrote:

jakolik wrote:

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\)

Ways to chose 6 members committee with two particular men serve together: \(C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65\)

700-65 = 635

Answer: E.

Committee has following formations- M----------W 2----------4 3----------3

Quote:

Ways to chose 6 members committee without restriction (two men refuse to server together): \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\)

This value 700 includes committees in which both of the men (say X and Y) are included. So, we need to find out number of such committees and deduct those from the total 700.

In the first case, 2 men are chosen. No. of committees of 2 men in which X and Y are included is 1 So, Considering women also we find total such committees\(= 1 * ^5C_4\) \(= 5\)

In the second case, 3 men are chosen out of 6. Also, out of 3 men X and Y will always be in the committee. Therefore, we need to find out 1 more man out of remaining 6. This can be done \(^6C_1\)way = 6 ways

So, Considering women also we find total such committees\(= 6 * ^5C_3\) \(= 6*10\)

Therefore, total no. of committees in which both X and Y are included is 5+60=65

This gives us answer, total no. of committees in which both X and Y are not included\(= 700-65=635\) _________________

My mantra for cracking GMAT: Everyone has inborn talent, however those who complement it with hard work we call them 'talented'.

+1 Kudos = Thank You Dear Are you saying thank you?

Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
29 Mar 2015, 07:46

1

This post received KUDOS

Expert's post

jakolik wrote:

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

Let M = men and W = women A committee of 6 is to be chosen. At least 2M and 3W should be there. So we can have either 2M+4W OR 3M+3W.

Total possibilities = (8C2)*(5C4) + (8C3)*(5C3) = 700.

But we are told that two men refuse to serve together.

For, 2M+4W we can select 2M from only 2 men because two men refuse to serve together. So, (2C2)*(5C4) = 5.

For, 3M+3W we can select 2M from only 2 men because two men refuse to serve together and the other 1M can be selected from remaining 6M. = 2M + 1M + 3W = (2C2)*(6C1)*(5C3) = 60. Hence total = 5 + 60 = 65.

Re: Another Combination Problem [#permalink]
29 Jul 2010, 20:33

Thank you Bunuel.

Can you please explain why ways to chose 6 members committee without restriction cannot be: C^2_8*C^3_5*C^1_8 We are to choose 2 men out of 8, 3 women out of 5 and 1 (woman or man) out of 8 (13-5)

Re: Another Combination Problem [#permalink]
29 Jul 2010, 21:05

jakolik wrote:

Thank you Bunuel.

Can you please explain why ways to chose 6 members committee without restriction cannot be: C^2_8*C^3_5*C^1_8 We are to choose 2 men out of 8, 3 women out of 5 and 1 (woman or man) out of 8 (13-5)

This way there will be duplicates in the combinations.

Consider M1,M2....M8 be men and W1,W2......W5 be women See the below 3 combinations. They are in fact same but will be counted as different with the approach you mentioned

Re: Another Combination Problem [#permalink]
17 Oct 2013, 09:19

Bunuel wrote:

jakolik wrote:

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\)

Ways to chose 6 members committee with two particular men serve together: \(C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65\)

700-65 = 635

Answer: E.

Hi, I went at it a different way and did not get the right result. Can you find out where the logic is wrong?

Total number of possibilities: choose 2 men, choose 3 women, choose one from the leftover: 8c2*5c3*8c1 Total number of possibilities for the 2 men together: 2C2*5C3*8C1. And last step, subtract the two numbers.... This is not working for me... Why?

Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
17 Oct 2013, 09:44

jakolik wrote:

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

Bunuel's subtraction way works. Here's another method -

There are 2 men and 3 women leaving out one spot.

Case 1: We have 3 men and 3 women. Total number of ways = (8C3 - 6)* 5C3 = 500 Case 2: We have 2 men and 4 women. Total number of ways = (8C2 - 1)*5C4 = 135

Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
18 Oct 2013, 02:06

jeopardo wrote:

jakolik wrote:

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

Bunuel's subtraction way works. Here's another method -

There are 2 men and 3 women leaving out one spot.

Case 1: We have 3 men and 3 women. Total number of ways = (8C3 - 6)* 5C3 = 500 Case 2: We have 2 men and 4 women. Total number of ways = (8C2 - 1)*5C4 = 135

Total = 635

I understand that Bunuels way works.... My question is why my way doesn't?

Re: Another Combination Problem [#permalink]
20 Oct 2013, 12:23

Expert's post

ronr34 wrote:

Bunuel wrote:

jakolik wrote:

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\)

Ways to chose 6 members committee with two particular men serve together: \(C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65\)

700-65 = 635

Answer: E.

Hi, I went at it a different way and did not get the right result. Can you find out where the logic is wrong?

Total number of possibilities: choose 2 men, choose 3 women, choose one from the leftover: 8c2*5c3*8c1 Total number of possibilities for the 2 men together: 2C2*5C3*8C1. And last step, subtract the two numbers.... This is not working for me... Why?

Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
10 Nov 2014, 06:50

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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
28 Mar 2015, 10:39

vomhorizon wrote:

Find out how many different combinations can be made between the 8 men and 5 women ...

We will have two scenarios , First where we have 2 men and 4 women , and second where we have 3 men and 4 women ... Finding the combinations for each and then adding them up gives us 700 ...

Logically the number of combinations where any 2 men do not serve together would have to be lower than 700 , therefore the only option is E ....

I didn't see this until I answered the question, but this is a quicker way to get to the answer.

gmatclubot

Re: A committee of 6 is chosen from 8 men and 5 women so as to
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28 Mar 2015, 10:39

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