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# A committee of 6 is chosen from 8 men and 5 women so as to contain at

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A committee of 6 is chosen from 8 men and 5 women so as to contain at  [#permalink]

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02 Sep 2010, 03:00
1
00:00

Difficulty:

45% (medium)

Question Stats:

71% (02:19) correct 29% (02:54) wrong based on 136 sessions

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A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed?

A. 635
B. 700
C. 1404
D. 2620
E. 3510

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Re: A committee of 6 is chosen from 8 men and 5 women so as to contain at  [#permalink]

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02 Sep 2010, 03:10
1
rafi wrote:
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed?

(1) 635
(2) 700
(3) 1404
(4) 2620
(5) 3510

I have no idea how to solve this question, every thing I come up with is not one of the possible answers..
Thanks..

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

So # of committees equals to: $$C^2_8*C^4_5+C^3_8*C^3_5 = 700$$.

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Re: A committee of 6 is chosen from 8 men and 5 women so as to contain at  [#permalink]

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12 Jul 2015, 13:03
1
reto wrote:
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed?

A. 635
B. 700
C. 1404
D. 2620
E. 3510

The only cases possible are :

1. 2 men and 4 women : 8C2 * 5C4 = 140

2. 3 men and 3 women: 8C3*5C3 = 560

Rest of the cases will either have 1 or 0 men (not allowed) or will have 1 or 2 or 0 women (not allowed)

Total possible combinations = 140+560 = 700. Thus B is the correct answer.
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Re: A committee of 6 is chosen from 8 men and 5 women so as to contain at  [#permalink]

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20 Aug 2015, 22:02
1
ggurface wrote:
Why is it not correct to do

8c2 * 5c3 * 8

where 8 is the remaining 6 men + 2 women?

Because you are double counting here.

When you select seperately but from the same group and do not arrange, there are chances of double counting. What that means is this:

Say, the 8 men are M1, M2, M3... M8.
Say, the 5 women are W1, W2, ... W5.

Case 1:
When you do 8C2, you select M1 and M2.
When you do 5C3, you select W1, W2 and W3
When you do 8C1 from the leftovers, you select M3
The group of 6 is M1, M2, M3, W1, W2, W3

Case 2:
When you do 8C2, you select M2 and M3.
When you do 5C3, you select W1, W2 and W3
When you do 8C1 from the leftovers, you select M1
The group of 6 is M1, M2, M3, W1, W2, W3

Though you counted these two cases as 2, they are actually the same because they give you the same group of people. The issue came up because you selected 2 men and then the 1 man separately from the same group on men. There was no arrangement required so it led to double counting cases. Hence, when you select from one group, you do it one go if you don't have to arrange.
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Re: A committee of 6 is chosen from 8 men and 5 women so as to contain at  [#permalink]

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20 Aug 2015, 18:35
Why is it not correct to do

8c2 * 5c3 * 8

where 8 is the remaining 6 men + 2 women?
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Re: A committee of 6 is chosen from 8 men and 5 women so as to contain at  [#permalink]

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09 Aug 2017, 15:17
we can have:
3 men 3 women
2 men 4 women

5C4 * 8C2 + 5C3 * 8C3 = 140 + 560 = 700.
Re: A committee of 6 is chosen from 8 men and 5 women so as to contain at   [#permalink] 09 Aug 2017, 15:17
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# A committee of 6 is chosen from 8 men and 5 women so as to contain at

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