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# A committee of 3 men and 3 women must be formed from a group of 6 men

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GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4163
A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

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02 Feb 2005, 19:11
3
28
00:00

Difficulty:

65% (hard)

Question Stats:

39% (02:21) correct 61% (02:20) wrong based on 88 sessions

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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?

(A) 1120
(B) 910
(C) 810
(D) 560
(E) 210

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Best Regards,

Paul

Math Expert
Joined: 02 Sep 2009
Posts: 52278
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

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02 Sep 2013, 00:50
6
6
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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Director
Joined: 19 Nov 2004
Posts: 525
Location: SF Bay Area, USA
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

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02 Feb 2005, 21:51
= Total combination - combination in which the man and women serve together
6c3*8c3 - 5c2*7c2
Intern
Joined: 28 Dec 2004
Posts: 32
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

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05 Mar 2005, 18:52
Hi,

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

Thanks,

Mike
Director
Joined: 18 Feb 2005
Posts: 639
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

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05 Mar 2005, 20:11
2
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
VP
Joined: 13 Jun 2004
Posts: 1074
Location: London, UK
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Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

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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
VP
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Location: Germany
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

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06 Mar 2005, 06:40
1
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).
Manager
Joined: 27 Jan 2005
Posts: 97
Location: San Jose,USA- India
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

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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 of 6 men  [#permalink]

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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
Math Expert
Joined: 02 Sep 2009
Posts: 52278
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

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02 Sep 2013, 00:53
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.

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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 of 6 men  [#permalink]

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 52278
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

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20 Apr 2014, 02:08
2
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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Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

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20 Apr 2018, 01:21
1
Paul wrote:
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?

Please find the solution with two methods in attachment

Bunuel

A) 1120
B) 910
C) 810
D) 560
E) 210
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Screen Shot 2018-04-20 at 2.48.06 PM.png [ 447.24 KiB | Viewed 6023 times ]

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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 2018, 01:23
GMATinsight wrote:
Paul wrote:
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?

Please find the solution with two methods in attachment

Bunuel

A) 1120
B) 910
C) 810
D) 560
E) 210

_______________
Done. Thank you.
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Re: A committee of 3 men and 3 women must be formed from a group &nbs [#permalink] 20 Apr 2018, 01:23
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