The number of ways in which $10$ candidates $A 1, A 2, A 3, A 4 . . . . A 10$ can be ranked if $A 1$ is just above $A 2$ , then the number of ways are

9! 2!

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10!

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10! 2!

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9!

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Solution

The correct option is D $9!$
Once again taking $A_{1}$ $A_{2}$ as a single block,
$A_{1} A_{2}$ $A_{3}$ $A_{4}$ $A_{5}$ $A_{6}$ $A_{7}$ $A_{8}$ $A_{9}$ $A_{10}$
We have $9$ blocks to rank, hence,
the no. of ways $= 9!$
However, $A_{1} A_{2}$ can not be switched this time, as $A_{1}$ has to be ranked just above $A_{2}$ .
So, the no. of ways in which the block rearranges itself is only $1!$
So. the total no. of ways $= 9!$

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