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# 3 men and 3 women need to be seated in 2 rows with 3 chairs

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Director
Joined: 07 Jun 2004
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3 men and 3 women need to be seated in 2 rows with 3 chairs [#permalink]

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28 Sep 2010, 05:47
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Difficulty:

25% (medium)

Question Stats:

68% (00:29) correct 32% (00:33) wrong based on 71 sessions

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3 men and 3 women need to be seated in 2 rows with 3 chairs in each row. All men have to be seated in the back row . How many ways can this be done.

A. 6
B. 12
C. 24
D. 36
E. 72
[Reveal] Spoiler: OA

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Kudos [?]: 969 [0], given: 22

Intern
Joined: 30 Mar 2010
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Re: Counting [#permalink]

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28 Sep 2010, 06:01
_ _ _ = 3! ways women can be seated
_ _ _ = 3! ways men can be seated
3! x 3!= 36 ways total
Does anyone else agree?

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Director
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Re: Counting [#permalink]

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28 Sep 2010, 06:10
I agree

what i had a doubt was since men have to be in the back row thats 3! ways and because of this women will be in the front row thats 3! why cant it be 6 + 6 = 12 ways since the row restriction is already in place
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Re: Counting [#permalink]

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28 Sep 2010, 06:22
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rxs0005 wrote:
I agree

what i had a doubt was since men have to be in the back row thats 3! ways and because of this women will be in the front row thats 3! why cant it be 6 + 6 = 12 ways since the row restriction is already in place

Principle of Multiplication
If one event can occur in $$m$$ ways and a second can occur independently of the first in $$n$$ ways, then the two events can occur in $$mn$$ ways.

Or consider this: for one particular arrangement of men, say {m1, m2, m3} women in the front row can be arranged in 3!=6 ways, as total # of arrangements of men is 3!=6 then total # of arrangements of men and women is 3!*3!=36.

Hope it's clear.
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Re: Counting [#permalink]

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28 Sep 2010, 07:09
Front: 3! = 6

Back: 3! = 6

6*6=36

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Re: 3 men and 3 women need to be seated in 2 rows with 3 chairs [#permalink]

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22 Aug 2014, 03:13
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Re: 3 men and 3 women need to be seated in 2 rows with 3 chairs   [#permalink] 22 Aug 2014, 03:13
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# 3 men and 3 women need to be seated in 2 rows with 3 chairs

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