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# Sits and Row

Author Message
Manager
Joined: 29 May 2008
Posts: 113
Followers: 1

Kudos [?]: 61 [0], given: 0

Sits and Row [#permalink]  17 Jun 2009, 05:25
00:00

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 1 sessions
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many ways can three people sit in a row of five sits if two of the five seats will remain empty?

A) 60
B) 30
C) 20
D) 15
E) 3
Senior Manager
Joined: 08 Jan 2009
Posts: 329
Followers: 2

Kudos [?]: 115 [0], given: 5

Re: Sits and Row [#permalink]  17 Jun 2009, 05:43
Three people can select 5 places in 5C3 ways and the can be rotated among themselves in 3! ways

5C3 * 3! = 10 * 6 = 60.
Manager
Joined: 29 May 2008
Posts: 113
Followers: 1

Kudos [?]: 61 [0], given: 0

Re: Sits and Row [#permalink]  17 Jun 2009, 05:59
It is soooo true those 3 have to stick together "emtpy"

Thank you so much
Intern
Joined: 24 May 2009
Posts: 38
Followers: 0

Kudos [?]: 28 [0], given: 4

Re: Sits and Row [#permalink]  19 Jun 2009, 18:33
Let $$P_1$$ ,$$P_2$$ , $$P_3$$ = 3 distinct persons
E = an empty seat

Therefore, there are $$P_1$$, $$P_2$$, $$P_3$$, E, E members to be rotated. They're 5! methods to rotate them.

However, there're 2 Es in common. So the answer is $$\frac{5!}{2!}$$ = 60. Agree with tkarthi4u, (A).
Re: Sits and Row   [#permalink] 19 Jun 2009, 18:33
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