Challenge : 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 are three persons ahead of C and 21 persons behind A, what could be the minimum number of persons in the queue? Any answer which I think is copied will not be marked brainliest and will be reported. Hint : Answer is not 40.
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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. Now, 28 < 40.
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