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A committee of 3 men and 3 women must be formed from a group

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A committee of 3 men and 3 women must be formed from a group [#permalink] New post 02 Feb 2005, 19:11
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A committee of 3 men and 3 women must be formed from a group of 6 men and 8 women. How many such committees can we form if 1 man and 1 woman refuse to serve together?
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 [#permalink] New post 02 Feb 2005, 21:51
= Total combination - combination in which the man and women serve together
6c3*8c3 - 5c2*7c2
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 [#permalink] New post 02 Feb 2005, 22:01
6C3x8C3 - 5C2x7C2 = 910
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 [#permalink] New post 03 Feb 2005, 07:20
:yes 910 is the OA. Too simple it is :lol:
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confused [#permalink] New post 05 Mar 2005, 18:52
Hi,

I was a little confused about how you determine: (5,2)*(7,2)?

Thanks,

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 [#permalink] New post 05 Mar 2005, 19:29
910

no of ways= total no of ways- when they are always together
=6c3*8c3-5c2*7c2
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not sure [#permalink] New post 05 Mar 2005, 19:33
sorry.. i'm not sure how everyone arrived at " 5c2*7c2 "

why does this equal the number of combinations that the 1 man and 1 women who can't be together? :oops:
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 [#permalink] New post 05 Mar 2005, 20:11
1 man and 1 woman are already selected so
You can select the remaining 2 men and 2 women from 5 men and 7 women

So 5C2*7C2
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 [#permalink] New post 06 Mar 2005, 02:25
WOOOOOOOOOOOOOOOOOO
910!!!! That was written on my paper!
Am I dreaming?
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 [#permalink] New post 06 Mar 2005, 06:37
gmat2me2 wrote:
1 man and 1 woman are already selected so
You can select the remaining 2 men and 2 women from 5 men and 7 women

So 5C2*7C2


sorry guys, I don't get it.
I agree on the way to calculate it : total outcome - outcome when the man and the woman are together in the group
I found 6C3*8C3 - 6C1*8C1..which is wrong but i can not understand why my answer is wrong and why 5c2*7c2 is good ? 5c2*7c2 just deal with 2 people , it seems incomplete to me... please help
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 [#permalink] New post 06 Mar 2005, 06:40
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Antmavel wrote:
gmat2me2 wrote:
1 man and 1 woman are already selected so
You can select the remaining 2 men and 2 women from 5 men and 7 women

So 5C2*7C2


sorry guys, I don't get it.
I agree on the way to calculate it : total outcome - outcome when the man and the woman are together in the group
I found 6C3*8C3 - 6C1*8C1..which is wrong but i can not understand why my answer is wrong and why 5c2*7c2 is good ? 5c2*7c2 just deal with 2 people , it seems incomplete to me... please help


5c2*7c2 => means that THE women and THE man is already in the group. so 4 places are left for 2 out of 5 (5c2) and 2 out of 7 (7c2).
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thanks!! [#permalink] New post 06 Mar 2005, 06:43
Thanks! I get it now!! I should have got it before. :shock:
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 [#permalink] New post 08 Mar 2005, 23:17
Antmavel wrote:
I found 6C3*8C3 - 6C1*8C1..which is wrong but i can not understand why my answer is wrong and why 5c2*7c2 is good ? 5c2*7c2 just deal with 2 people , it seems incomplete to me... please help


Question says 1 man and 1 woman refuse to server. Let's say persons refuse to server are John and Mary. 6C1*8C1 would mean you are selecting any 1 man from 6 men(John may or may not be there in the selection) and any 1 woman from 8 women(Mary may or may not be there in the selection).So it is wrong.

Instead, we know John is already selected, so we are left with picking 2 men from 5,and as Mary is already selected we are left with picking 2 women from 7.Thus comes 5C2*7C2.
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Re: A committee of 3 men and 3 women must be formed from a group [#permalink] New post 01 Sep 2013, 21:15
Hi Bunuel,

Can u explain this problem

in the above explanation its states that "Question says 1 man and 1 woman refuse to server. Let's say persons refuse to server are John and Mary. 6C1*8C1 would mean you are selecting any 1 man from 6 men(John may or may not be there in the selection) and any 1 woman from 8 women(Mary may or may not be there in the selection).So it is wrong.

Instead, we know John is already selected, so we are left with picking 2 men from 5,and as Mary is already selected we are left with picking 2 women from 7.Thus comes 5C2*7C2. "

If John and mary refused to work together then how can we select those people already and subtract 5C2*7C2 from total combination???

Pls explain this.

Thanks and Regards,
Rrsnathan
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Re: A committee of 3 men and 3 women must be formed from a group [#permalink] New post 02 Sep 2013, 00:50
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rrsnathan wrote:
Hi Bunuel,

Can u explain this problem

in the above explanation its states that "Question says 1 man and 1 woman refuse to server. Let's say persons refuse to server are John and Mary. 6C1*8C1 would mean you are selecting any 1 man from 6 men(John may or may not be there in the selection) and any 1 woman from 8 women(Mary may or may not be there in the selection).So it is wrong.

Instead, we know John is already selected, so we are left with picking 2 men from 5,and as Mary is already selected we are left with picking 2 women from 7.Thus comes 5C2*7C2. "

If John and mary refused to work together then how can we select those people already and subtract 5C2*7C2 from total combination???

Pls explain this.

Thanks and Regards,
Rrsnathan


A committee of 3 men and 3 women must be formed from a group of 6 men and 8 women. How many such committees can we form if 1 man and 1 woman refuse to serve together?

The total # of committees without the restriction is C^3_6*C^3_8;
The # of committees which have both John and Mary is 1*1*C^2_5*C^2_7 (one way to select John, 1 way to select Mary, selecting the remaining 2 men from 5, selecting the remaining 2 women from 7).

{Total} - {Restriction} = C^3_6*C^3_8-C^2_5*C^2_7.

Hope it's clear.
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Re: A committee of 3 men and 3 women must be formed from a group [#permalink] New post 02 Sep 2013, 00:53
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Bunuel wrote:
rrsnathan wrote:
Hi Bunuel,

Can u explain this problem

in the above explanation its states that "Question says 1 man and 1 woman refuse to server. Let's say persons refuse to server are John and Mary. 6C1*8C1 would mean you are selecting any 1 man from 6 men(John may or may not be there in the selection) and any 1 woman from 8 women(Mary may or may not be there in the selection).So it is wrong.

Instead, we know John is already selected, so we are left with picking 2 men from 5,and as Mary is already selected we are left with picking 2 women from 7.Thus comes 5C2*7C2. "

If John and mary refused to work together then how can we select those people already and subtract 5C2*7C2 from total combination???

Pls explain this.

Thanks and Regards,
Rrsnathan


A committee of 3 men and 3 women must be formed from a group of 6 men and 8 women. How many such committees can we form if 1 man and 1 woman refuse to serve together?

The total # of committees without the restriction is C^3_6*C^3_8;
The # of committees which have both John and Mary is 1*1*C^2_5*C^2_7 (one way to select John, 1 way to select Mary, selecting the remaining 2 men from 5, selecting the remaining 2 women from 7).

{Total} - {Restriction} = C^3_6*C^3_8-C^2_5*C^2_7.

Hope it's clear.


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Re: A committee of 3 men and 3 women must be formed from a group [#permalink] New post 19 Apr 2014, 18:48
Hi,

I tried to do it the other way around, instead of: total combos - combos restricted, I tried the approach of adding up all permited combos.

Combos where the man is but the woman is left out: 6C3*7C3 (Only take 7 women into account, not 8)
Combos where the woman is but the man is left out: 5C3*8C3 (Only take 5 men into account, not 6)
Combos where neither is in a group selected: 5C3*7C3 (Both are taken out)

Adding up these three scenarios, I get a total of 1,530 combos.

Could someone help me out here? Can't seem to understand where i'm over estimating.

Much appreciated.
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Re: A committee of 3 men and 3 women must be formed from a group [#permalink] New post 20 Apr 2014, 02:08
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Enael wrote:
Hi,

I tried to do it the other way around, instead of: total combos - combos restricted, I tried the approach of adding up all permited combos.

Combos where the man is but the woman is left out: 6C3*7C3 (Only take 7 women into account, not 8)
Combos where the woman is but the man is left out: 5C3*8C3 (Only take 5 men into account, not 6)

Combos where neither is in a group selected: 5C3*7C3 (Both are taken out)

Adding up these three scenarios, I get a total of 1,530 combos.

Could someone help me out here? Can't seem to understand where i'm over estimating.

Much appreciated.


The number of committees with John but not Mary: (1*C^2_5)(C^3_7)=10*35=350;

The number of committees with Mary but not John: (C^3_5)(1*C^2_7)=10*21=210;

The number of committees without John and without Mary: (C^3_5)(C^3_7)=10*35=350.

Total = 350 + 350 + 210 = 910.

Hope it's clear.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: A committee of 3 men and 3 women must be formed from a group   [#permalink] 20 Apr 2014, 02:08
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