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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
[Reveal] Spoiler: OA
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Re: Another Combination Problem [#permalink]

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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.
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Re: Another Combination Problem [#permalink]

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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.
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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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 ....
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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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\(= 700-65=635\)
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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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.

Now, required possibilities = 700 - 65 = 635
Hence option (E).

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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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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


First case 2 men and 4 Women

Let M1 and M2 refuse to serve together

When M1 serves and not M2 , number of ways 6C1*5C4=30
Similarly 30 ways when M2 and not M1 serves
When both M1 nd M2 do not serve 6C2*5C4=75
Total number of ways = 135

Second case of 3 men and 3 women

Using the above reasoning number of ways is 6C2*5C3 + 6C2*5C3 + 6C3*5C3 = 500

Total = 135+500=635
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Re: Another Combination Problem [#permalink]

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New post 29 Jul 2010, 21:33
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)
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Re: Another Combination Problem [#permalink]

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New post 29 Jul 2010, 22:05
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

...2M.. .....3W...... 1M or 1W
M1,M2| W1,W2,W3| M3
M3,M2| W1,W2,W3| M1
M3,M1| W1,W2,W3| M2

Thanks
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Re: Another Combination Problem [#permalink]

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New post 31 Jul 2010, 09:41
Quote:
Ways to chose 6 members committee with two particular men serve together: \(C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65\)


Bunuel, can you explain the logic behind this?
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Re: Another Combination Problem [#permalink]

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New post 15 Sep 2012, 07:04
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 ?
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Re: Another Combination Problem [#permalink]

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New post 17 Oct 2013, 10:19
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?
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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New post 17 Oct 2013, 10:44
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

Total = 635
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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New post 18 Oct 2013, 03:06
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?
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Re: Another Combination Problem [#permalink]

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New post 20 Oct 2013, 13:23
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?


Check here: 4-professors-and-6-students-are-being-considered-for-85865-20.html#p1265096
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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New post 28 Mar 2015, 11:39
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.
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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New post 10 Jan 2016, 19:21
spent some time solving this bad boy..
so we can have 2 options:
2M and 4W
and
3M and 3W.

8C2*5C4 + 8C3*5C3 = 700.
this is without restrictions. logically, we can eliminate all answer choices except for E, which is less than 700. and it has to be 700,if we apply restrictions.

nevertheless, with restrictions, I found it difficult to solve...
tried to combine the 2 guys as 1 person:
7C1 * 5C4 = 35
7C2 * 5C3 = 210
245 total ways..but this is way too much...
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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New post 26 Jan 2016, 03:49
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 Brunel, I understood process of finding solution with total-unaccepted solution

But before looking your explanation I started solving question in below manner-


1)2M and 4W
we have restriction that both person should not be together,one of them can be appear in commitee though.

so lets say A & B shouldnt be together
a)when one of them is present lets say A is present and B is not we have 7C2*5C4
b)when both A and B not selected we have option as 6C2*5C4
a+b=total ways of forming committee=180

2)3M3W

a)when one of them is present lets say A is present and B is not we have 7C3*5C3
b)when both A and B not selected we have option as 6C3*5C3
a+b=total ways of forming committee=550

so TOTAL=180+550=730

let me know please which combination I included twice..why I am getting wrong ans

Thanks in advance
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A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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New post 26 Jan 2016, 06:09
Expert's post
Hi,
the method you have adopted is longer and prone to error..
It is much simpler to find ways both are together and subtract from total to get the answer..
lets see the error in your solution..

Quote:
1)2M and 4W
we have restriction that both person should not be together,one of them can be appear in commitee though.

so lets say A & B shouldnt be together
a)when one of them is present lets say A is present and B is not we have 7C2*5C4
b)when both A and B not selected we have option as 6C2*5C4
a+b=total ways of forming committee=180

the error is in coloured portion..
1) 7C2 means selecting 2 out of 7, so this is including ways where even A is not included..
2) It just means all the way where B is not there..
3) So your equation where A is surely there and B is not there will be 6C1*5C4, as one of the men is already occupied by A = 30,..
4) the answer of 3) should be doubled, as it would give us where B is there and A is not there =30*2=60..
5) total ways would be 6C2*5C4+60=135..

Quote:
2)3M3W

a)when one of them is present lets say A is present and B is not we have 7C3*5C3
b)when both A and B not selected we have option as 6C3*5C3
a+b=total ways of forming committee=550

so TOTAL=180+550=730


I'll leave it for you to do in the similar way as the ways for 2M4W..
it should get you 500


Quote:
let me know please which combination I included twice..why I am getting wrong ans

Thanks in advance

hope you are clear where you went wrong..
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

A committee of 6 is chosen from 8 men and 5 women so as to   [#permalink] 26 Jan 2016, 06:09

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