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In a school show there were 5 identical chairs in the first row for VI

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In a school show there were 5 identical chairs in the first row for VI  [#permalink]

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New post Updated on: 18 Nov 2018, 02:29
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Question Stats:

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In a school show there were 5 identical chairs in the first row for VIPs, of which 2 were kept for the ministers, 1 for the principal and 2 for the chief guests. How many seating arrangements are possible in the first row if both the ministers were absent?

A. 30
B. 40
C. 60
D. 120
E. 600

Originally posted by NehaW on 18 Nov 2018, 02:26.
Last edited by Bunuel on 18 Nov 2018, 02:29, edited 1 time in total.
Renamed the topic and edited the question.
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Re: In a school show there were 5 identical chairs in the first row for VI  [#permalink]

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New post 18 Nov 2018, 07:42
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OA:C

Total number Seating arrangement that are possible in the first row if both the ministers were absent =5*4*3 =60
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Re: In a school show there were 5 identical chairs in the first row for VI  [#permalink]

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New post 20 Jun 2019, 08:25
NehaW wrote:
In a school show there were 5 identical chairs in the first row for VIPs, of which 2 were kept for the ministers, 1 for the principal and 2 for the chief guests. How many seating arrangements are possible in the first row if both the ministers were absent?

A. 30
B. 40
C. 60
D. 120
E. 600


Bunuel, Could you please explain this one.
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In a school show there were 5 identical chairs in the first row for VI  [#permalink]

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New post 20 Jun 2019, 15:08
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Its a combinations problem with redundant elements. The redundant or repeating elements are the two chief guests, belonging in the denominator. The two ministers are not included in the denominator because they are absent.

5! 5 x 4 x 3 x 2 x 1
__ = ______________ = 5 x 4 x 3 = 60.
2! 2 x 1

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Re: In a school show there were 5 identical chairs in the first row for VI  [#permalink]

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New post 01 Aug 2019, 23:05
gmat2019now wrote:
Its a combinations problem with redundant elements. The redundant or repeating elements are the two chief guests, belonging in the denominator. The two ministers are not included in the denominator because they are absent.

5! 5 x 4 x 3 x 2 x 1
__ = ______________ = 5 x 4 x 3 = 60.
2! 2 x 1

Posted from my mobile device


Hi,

I dont think the reasoning you gave is logical .
As per me the explanation is
NOW we have 5 seats for 2 chief ministers and 1 president so
5!/2! where 5! is for total seats and 2! is for identical Chief Ministers . answer is 5x4x3x2x1 / 2x1 = 60
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Re: In a school show there were 5 identical chairs in the first row for VI  [#permalink]

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New post 07 Aug 2019, 04:02
We have 3 people to occupy 5 chairs. 1 st person has 5 choices, 2nd person has 4 choices and 3rd person has 3 choices making total possible permutations equal to 5x4x3= 60
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Re: In a school show there were 5 identical chairs in the first row for VI   [#permalink] 07 Aug 2019, 04:02
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In a school show there were 5 identical chairs in the first row for VI

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