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

Question

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

Bunuel · Accepted Answer

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.

nocilis · Answer

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

mlpxmlp · Answer

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

gmat2me2 · Answer

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

Antmavel · Answer

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

christoph · Answer

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).

700Plus · Answer

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.

rrsnathan · Answer

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

Bunuel · Answer

Similar questions: at-a-meeting-of-the-7-joint-chiefs-of-staff-the-chief-of-154205.html a-committee-of-6-is-chosen-from-8-men-and-5-women-so-as-to-104859.html

Enael · Answer

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.

Bunuel · Answer

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.

GMATinsight · Answer

Please find the solution with two methods in attachment

Bunuel

Could you please add the options in question???

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

Bunuel · Answer

_______________
Done. Thank you.

bumpbot · Answer

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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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