Question

Three people M, N and P are standing in a queue. Five people are standing between M and N and eight people are standing between N and P. If there are three people ahead of P and 21 people behind M, what could be the minimum number of people in the queue?

41

40

27

None of these

28

Solution

The correct option is **E**

28

Three people M, N, P can be arranged in a queue in six different ways, ie MNP, PNM, NMP, PMN, NPM, MPN. But since there are only 3 people ahead of P, so P should be in front of the queue. Thus, there are only two possible arrangements, ie PNM and PMN.

We may consider the two cases as under:

**Case I:**

(3 people) P (8 people) N (5 people) M (21 people)

Clearly, number of people in the queue = 3+1+8+1+5+1+21= 40

**Case II: **

(3 people) P M (5 people) N

Number of people between M and P

= (8 - 6) = 2 [Since P (8 people) N, M (21 people)]

Clearly number of people in the queue = (3+1+2+1+21) = 28

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

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