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. $41$
• B. $40$
• C. $28$
• D. $27$

Answer 1

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 $28.$