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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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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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New post 02 Feb 2005, 22:51
= 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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New post 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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New post 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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New post 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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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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New post 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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New post 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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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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Re: A committee of 3 men and 3 women must be formed from a group [#permalink]

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New post 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.


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

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New post 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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New post 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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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