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

Hi,

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

Thanks,

Mike

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).

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.

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

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

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.

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

_______________
Done. Thank you.

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