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# The co-op board of a certain residential building must consist of two

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Math Expert
Joined: 02 Sep 2009
Posts: 65062
The co-op board of a certain residential building must consist of two  [#permalink]

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30 Apr 2019, 23:14
00:00

Difficulty:

15% (low)

Question Stats:

84% (01:29) correct 16% (01:53) wrong based on 68 sessions

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The co-op board of a certain residential building must consist of two men and three women. If there are six men and seven women who want to be on the committee, how many different make-ups of the committee exist?

A. 65
B. 525
C. 1,050
D. 1,287
E. 100,800

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Re: The co-op board of a certain residential building must consist of two  [#permalink]

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01 May 2019, 00:18
Bunuel wrote:
The co-op board of a certain residential building must consist of two men and three women. If there are six men and seven women who want to be on the committee, how many different make-ups of the committee exist?

A. 65
B. 525
C. 1,050
D. 1,287
E. 100,800

two men and three women to be selected out of six men and seven women

Total ways = 6C2 * 7C3 = 15 * 35 = 525

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Re: The co-op board of a certain residential building must consist of two  [#permalink]

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01 May 2019, 01:24
Bunuel wrote:
The co-op board of a certain residential building must consist of two men and three women. If there are six men and seven women who want to be on the committee, how many different make-ups of the committee exist?

A. 65
B. 525
C. 1,050
D. 1,287
E. 100,800

men ; 6c2 ; woment 7c3
solve 525
IMO B
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Re: The co-op board of a certain residential building must consist of two  [#permalink]

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29 Mar 2020, 03:53
Bunuel wrote:
The co-op board of a certain residential building must consist of two men and three women. If there are six men and seven women who want to be on the committee, how many different make-ups of the committee exist?

A. 65
B. 525
C. 1,050
D. 1,287
E. 100,800

6C2 x 7C3 = [(6 x 5) / 2] x [(7 x 6 x 5) / (3 x 2)] = 15 x 35 = 525

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Re: The co-op board of a certain residential building must consist of two  [#permalink]

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29 Mar 2020, 06:15

Solution

Given
In this question, we are given that
• The co-op board of a certain residential building must consist of two men and three women.
• There are six men and seven women who want to be on the committee.

To find
We need to determine
• The number of different committees that can exist

Approach and Working out
In the formation of the committee,
• 2 men need to be chosen from 6 men in $$^6C_2$$ or 15 ways
• 3 women need to be chosen from 7 women in $$^7C_3$$ or 35 ways

Hence, the number of ways the formation can be done = 15 x 35 = 525

Thus, option B is the correct answer.

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Re: The co-op board of a certain residential building must consist of two   [#permalink] 29 Mar 2020, 06:15