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Three persons A, B and C are standing in a queue. There are five persons between A and B and eight persons between B and C. If there be three persons ahead of C and 21 persons behind A, what could be the minimum number of persons in the queue?

Correct Answer: 28

Three persons A, B, C can be arranged in a queue in six different ways, ie ABC, CBA, BAC, CAB, BCA, ACB. But since there are only 3 persons ahead of C, so C should be in front of the queue. Thus, there are only two possible arrangements, ie CBA and CAB.

 

We may consider the two cases as under:

Case I:  ←3C↔8B↔5A→21

 

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

 

Case II:  ←3C A↔5B

 

Number of persons between A and C                  

 

= (8 - 6) = 2   

∵[C↔8B A→21B]

 

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.                                                     

 

 

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