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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]

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02 Feb 2005, 20:11
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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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Paul

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Re: A committee of 3 men and 3 women must be formed from a group [#permalink]

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02 Sep 2013, 01:50
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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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05 Mar 2005, 21:11
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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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Re: A committee of 3 men and 3 women must be formed from a group [#permalink]

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20 Apr 2014, 03:08
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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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06 Mar 2005, 03:25
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WOOOOOOOOOOOOOOOOOO
910!!!! That was written on my paper!
Am I dreaming?
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06 Mar 2005, 07:40
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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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02 Feb 2005, 22:51
= Total combination - combination in which the man and women serve together
6c3*8c3 - 5c2*7c2
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02 Feb 2005, 23:01
6C3x8C3 - 5C2x7C2 = 910
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03 Feb 2005, 08:20
910 is the OA. Too simple it is
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Paul

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05 Mar 2005, 19:52
Hi,

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

Thanks,

Mike
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05 Mar 2005, 20:29
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910

no of ways= total no of ways- when they are always together
=6c3*8c3-5c2*7c2
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05 Mar 2005, 20:33
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?
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06 Mar 2005, 07:37
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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06 Mar 2005, 07:43
Thanks! I get it now!! I should have got it before.
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09 Mar 2005, 00:17
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]

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01 Sep 2013, 22:15
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
Math Expert
Joined: 02 Sep 2009
Posts: 38887
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Kudos [?]: 106130 [0], given: 11607

Re: A committee of 3 men and 3 women must be formed from a group [#permalink]

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02 Sep 2013, 01:53
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.

Similar questions:
at-a-meeting-of-the-7-joint-chiefs-of-staff-the-chief-of-154205.html
a-committee-of-6-is-chosen-from-8-men-and-5-women-so-as-to-104859.html
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Re: A committee of 3 men and 3 women must be formed from a group [#permalink]

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19 Apr 2014, 19:48
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]

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Re: A committee of 3 men and 3 women must be formed from a group [#permalink]

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26 May 2016, 21:58
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