Multiple choice

The number of ways in which the candidates $A1, A2, ..... A10$ can be arranged if $A1$ and $A2$ should always be together is

  1. $9! 2!$
  2. $9!$
  3. $\dfrac{10!}{2!}$
  4. $\dfrac{9!}{4!}$
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A Correct answer
Explanation

Treat A1 and A2 as a single unit. There are now 9 units to arrange (the block + 8 others), which is 9!. Within the block, A1 and A2 can be arranged in 2! ways. Total = 9! * 2!.

AI explanation

We treat the two candidates A1 and A2 as a single combined block since they must always be together. Including this block, there are now 9 distinct entities to arrange, which can be done in 9! ways. The candidates A1 and A2 can switch places within their block in 2! ways. Using the multiplication principle, the total number of arrangements is 9! x 2!.

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