Multiple choice

Fifteen friends are to be placed in a queue. Three of them always stand together and out of the remaining twelve, two do not want to stand together. Then, the number of possible arrangements is:

  1. P(12, 3) × 2!

  2. 12!3! × 11

  3. 12!3!2!

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Treat the 3 friends who stand together as one block. This leaves 13 units (1 block + 12 individuals). Total arrangements with the block is 13! * 3!. Subtract the cases where the 2 specific friends are together: treat them as a block, leaving 12 units, resulting in 12! * 3! * 2!. The difference is 13! * 3! - 12! * 3! * 2! = 12! * 3! * (13 - 2) = 12! * 3! * 11.

AI explanation

First, treat the 3 friends who stand together as a single unit, making 13 total units in the queue. The 12 individual friends can be arranged in 12! ways, the unit of 3 can be arranged internally in 3! ways, and these units can be arranged in 13! ways. To ensure the 2 specific friends among the 12 do not stand together, we subtract their together cases from the total, giving the formula 13!3! minus 12!2!3!2!, which factors to 12!3! multiplied by 11.

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