3 men and 3 women need to be seated in 2 rows with 3 chairs : Problem Solving (PS)

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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

Posts: 557

Location: PA

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

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28 Sep 2010, 06:47

00:00

Difficulty:

25% (medium)

Question Stats:

68% (00:57) correct

32% (00:53) wrong

based on 109 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

Spoiler: OA

Intern

Joined: 30 Mar 2010

Posts: 4

Re: Counting [#permalink]

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28 Sep 2010, 07:01

_ _ _ = 3! ways women can be seated
_ _ _ = 3! ways men can be seated
3! x 3!= 36 ways total
Does anyone else agree?

Director

Joined: 07 Jun 2004

Posts: 557

Location: PA

Re: Counting [#permalink]

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28 Sep 2010, 07:10

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

Math Expert

Joined: 02 Sep 2009

Posts: 58316

Re: Counting [#permalink]

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28 Sep 2010, 07:22

rxs0005 wrote:

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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Intern

Joined: 23 Apr 2009

Posts: 47

Location: Texas

Re: Counting [#permalink]

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28 Sep 2010, 08:09

Front: 3! = 6

Back: 3! = 6

6*6=36

Non-Human User

Joined: 09 Sep 2013

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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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16 Jan 2019, 21:40

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