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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, 20:11
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35% (01:58) correct 65% (01:45) wrong based on 77 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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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, 01:50
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). \({Total}  {Restriction} = C^3_6*C^3_8C^2_5*C^2_7\). Hope it's clear.
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Re: A committee of 3 men and 3 women must be formed from a group of 6 men [#permalink]
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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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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, 19:52
Hi,
I was a little confused about how you determine: (5,2)*(7,2)?
Thanks,
Mike



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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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05 Mar 2005, 21:11
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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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, 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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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, 07:40
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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Re: A committee of 3 men and 3 women must be formed from a group of 6 men [#permalink]
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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 of 6 men [#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



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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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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). \({Total}  {Restriction} = C^3_6*C^3_8C^2_5*C^2_7\). Hope it's clear. Similar questions: atameetingofthe7jointchiefsofstaffthechiefof154205.htmlacommitteeof6ischosenfrom8menand5womensoasto104859.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, 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 of 6 men [#permalink]
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20 Apr 2014, 03:08
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, 02:21
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 BunuelCould you please add the options in question??? A) 1120 B) 910 C) 810 D) 560 E) 210
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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, 02:23




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