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?	Choose one
A) 33
B) 37
C) 28
D) 36
E) 19
Explanation:   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:   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)= 
<gadget id="calculator">(8 - 6)</gadget>
<output>2</output>
2   Clearly number of persons in the queue= (3+1+2+1+21)= 
<gadget id="calculator">(3+1+2+1+21)</gadget>
<output>28</output>
28   Now, 28 &lt; 40. So, 28 is the minimum number of persons in the queue.   Answer: 28) 28
<result>C</result>
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?