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?

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:   $\leftarrow 3C\leftrightarrow 8B\leftrightarrow 5A\to 21$

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

Case II:   $\leftarrow 3C A\leftrightarrow 5B$

Number of persons between A and C                  

= (8 - 6) = 2   

$\because [C\leftrightarrow 8B A\to 21B]$

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.