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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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= Total combination  combination in which the man and women serve together
6c3*8c3  5c2*7c2



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confused [#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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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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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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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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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



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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). \({Total}  {Restriction} = C^3_6*C^3_8C^2_5*C^2_7\). Hope it's clear.
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Re: A committee of 3 men and 3 women must be formed from a group [#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 [#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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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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