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Question
Three persons A, B and C are standing in queue. There are five persons between A and B and eight persons between B and C. If there ne three persons ahead of C and 21 persons behind A, what could be the minimum number of persons in the queue ?
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Solution
The correct option is C 28
Three persons A, B, C can be arranged in a queue in six different ways i.e., 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 i.e., CBA and CAB. We may consider the two cases as under
3 8 5 21
Case I : ←←.
Now, 28 < 40. So, 28 is the minimum number of perosn in the queue.
Three persons A, B, C can be arranged in a queue in six different ways i.e., 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 i.e., CBA and CAB. We may consider the two cases as under
3 8 5 21
Case I : ←←.
Now, 28 < 40. So, 28 is the minimum number of perosn in the queue.
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