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A committee of 3 men and 3 women must be formed from a group [#permalink]
02 Feb 2005, 19:11

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Question Stats:

30% (02:02) correct
70% (02:11) wrong based on 16 sessions

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? _________________

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

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).

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

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.

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

Re: A committee of 3 men and 3 women must be formed from a group [#permalink]
02 Sep 2013, 00:53

Expert's post

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).

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

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

Expert's post

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: \((C^3_5)(C^3_7)=10*35=350\).

Re: A committee of 3 men and 3 women must be formed from a group [#permalink]
29 Apr 2015, 22:42

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