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

Question

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

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

omkartadsare · Accepted Answer

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

Bismarck · Answer

OA:C

Total number Seating arrangement that are possible in the first row if both the ministers were absent =5*4*3 =60

mzaid · Answer

Bunuel, Could you please explain this one.

gmat2019now · Answer

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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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

gmatway · Answer

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

mandeep2295 · Answer

The easiest way to solve this would be:

In case, ministers had come: To make 5 arrangements (M1, M2, P1, G1, G2) in 5 seats, the answer will be 5x4x3x2x1 = 120 ways (5 ways to fill 1st seat, then 4 ways to fill 2nd seat, then 3 ways to fill 3rd seat, then 2 ways to fill 4th seat & finally only 1 way/person left to fill 5th seat)

Now because the ministers are not coming: We have 3 people but we still need to make 5 arrangements (  Empty, Empty, P1, G1, G2  ) in 5 seats, the answer will be (5x4x3x2x1)/2! = 60 ways.

We divide by the factorial of the count of empty seats because we don't care about the order of the 2 empty seats & since Empty1, Empty2, P1, G1, G2 is the same as Empty2, Empty1, P1, G1, G2 & all such duplicated ways must be removed from the answer.

TestPrepUnlimited · Answer

This is a standard probability problem that should be mastered. There are 5 seats and 3 people, and it helps to think of the chairs as slots. The first person can take any of the 5 seats, after the first person is slotted the second person can take any of the remaining 4 seats, and for the third and final person he can take any of the remaining 3 seats. Therefore our total number of options are 5 * 4 * 3 = 60.

Note we should not repeat this process for different “first persons” because we already included all possible combinations.

Answer: C

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

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