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# If there are 4 seats and two students, in how many ways can

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Joined: 19 Nov 2013
Posts: 24
Location: India
Concentration: Strategy, Technology
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If there are 4 seats and two students, in how many ways can  [#permalink]

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Updated on: 22 Apr 2014, 10:56
5
00:00

Difficulty:

15% (low)

Question Stats:

70% (00:34) correct 30% (00:35) wrong based on 166 sessions

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If there are 4 seats and two students, in how many ways can they be seated?

A. 12
B. 6
C. 8
D. 16
E. 10

Originally posted by manish2014 on 22 Apr 2014, 10:44.
Last edited by Bunuel on 22 Apr 2014, 10:56, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Question on basic permutation  [#permalink]

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22 Apr 2014, 10:50
will it be 4 x 3 =12?

i.e 1st student can be seated in 4 ways
2nd student in 3 ways

therefore 4x3 = 12 ways?

can this be as 4p2?
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Posts: 56256
Re: Question on basic permutation  [#permalink]

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22 Apr 2014, 11:00
manish2014 wrote:
If there are 4 seats and two students, in how many ways can they be seated?

A. 12
B. 6
C. 8
D. 16
E. 10

will it be 4 x 3 =12?

i.e 1st student can be seated in 4 ways
2nd student in 3 ways

therefore 4x3 = 12 ways?

can this be as 4p2?

Both approaches give the same answer and both are correct.
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Re: If there are 4 seats and two students, in how many ways can  [#permalink]

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22 Apr 2014, 11:20
lets take another example : 5 red balls and 5 blacks. In how many ways can they be arranged so that b & w alternate?

soln--> _b_b_b_b_b_

so black can be arranged in 5! ways

and whites can take 6 positions '_'

there fore blacks take 6p5

so ans : 5! x 6p5?
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Posts: 56256
Re: If there are 4 seats and two students, in how many ways can  [#permalink]

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23 Apr 2014, 07:59
manish2014 wrote:
lets take another example : 5 red balls and 5 blacks. In how many ways can they be arranged so that b & w alternate?

soln--> _b_b_b_b_b_

so black can be arranged in 5! ways

and whites can take 6 positions '_'

there fore blacks take 6p5

so ans : 5! x 6p5?

If red balls are identical and black balls are identical too, then there can be only two arrangements: RBRBRBRBRB or BRBRBRBRBR.

If they are not identical, so if we are also concerned about the arrangements of red and black balls, then the answer would be 2*5!*5!.
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Re: If there are 4 seats and two students, in how many ways can  [#permalink]

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04 Nov 2015, 10:48
manish2014 wrote:
If there are 4 seats and two students, in how many ways can they be seated?

A. 12
B. 6
C. 8
D. 16
E. 10

Considering the Linear arrangement

No. of ways of choosing 2 chairs out of 4 chairs on which two students will sit = 4C2 = 6
No. of ways of 2 people sitting on selected chairs = 2!

Total ways of making 2 people sit on 2 of 4 chairs = 4C2*2! = 6*2 = 12

ALTERNATE

No. of ways of first person sitting on any one of the 4 chairs = 4 ways
No. of ways of Second person sitting on any one of the REMAINING 3 chairs = 3 ways

Total ways of making 2 people sit on 2 of 4 chairs = 4*3 = 12

NOTE: Since it's not mentioned in the question whether seating arrangemtn is Linear or circular so I find it a small weakness in the question because the circular arrangement will be different for same no. of people and chairs.
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Re: If there are 4 seats and two students, in how many ways can  [#permalink]

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22 Dec 2018, 15:49
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If there are 4 seats and two students, in how many ways can   [#permalink] 22 Dec 2018, 15:49
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