Question

Three persons A, B and C are standing in a queue. There are 5 persons between A and B and eight persons between B ad C. If there be 3 persons ahead of C and twenty one persons behind A, what could be the minimum number of persons in the queue?

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Answer 1

Answer:

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Step-by-step explanation:

Three persons A, B, C can be arranged in a queue in six different ways i.e., ABC, CBA, BAC, CAB, BCA, ACB. Butsince there are only 3 persons ahead of C, so C should be in front of the queue. Thus, there are only two possiblearrangements i.e., CBA and CAB.

We may consider the two cases as under :

3 8 5 21

Case I :C B A

Clearly, number of persons in the queue= (3 +1+8+1+5+1+21) = 40.

Case II :

Number of persons between A and C= (8 - 6) = 2.Clearly, number of persons in the queue= (3 + 1 + 2 + 1 + 21) = 28.Now, 28 <40. So, 28 is the minimum number of persons in the queue

Answer 2

Answer:

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