In Ranking Sequence questions, ranks of a person both from the top and bottom are mentioned and total number of persons is asked. However, sometimes this question is put in the form of a puzzle of interchanging seats by two persons.

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 ?

** Option:**

**A.**41

**B.**40

**C.**28

**D.**27

**Answer: C . 28**

** Justification:**

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 21Case I :** ← C ↔ B ↔ A →

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

** 3 5Case II :** ← C A ↔ B

Number of persons between A and C = (8 - 6) = 2.

Clearly, number of persons in the queue = (3 + 1 + 2 + 1 + 21) = 28.

Now, 28 < 40. So, 28 is the minimum number of perosn in the queue.