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Question

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


A
41
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B
40
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C
28
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D
27
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Solution

The correct option is C $$28$$
A, B, C can be arranged in a queue $$= 6! =$$ six different ways i.e., ABC, CBA, BAC, CAB, BCA, ACB. 
According to question, only $$3$$ persons can be ahead of C, 
So, there are only two possible arrangements i.e. CBA and CAB. 
Case 1) gives us $$40$$ as number of people in the queue.
 Number of persons $$=  3 + 1 + 8 + 1 + 5 + 1 + 21 = 40.$$
Case 2) gives us $$28$$ as number of people in the queue.
number of persons $$= (3 + 1 + 2 + 1 + 21) = 28.$$
So, minimum number is $$28.$$

524621_109183_ans_0a2117abba774948ad2bca8bcbdf417e.png

Logical Reasoning

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