The number of ways in which ten candidates A_ | vedclass

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(D) The total number of ways to rank $10$ candidates without any restrictions is $10!$.In any ranking,there are only two possibilities for the relativ

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$\frac{1}{2}(10!)$

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(D) The total number of ways to rank $10$ candidates without any restrictions is $10!$.In any ranking,there are only two possibilities for the relative positions of $A_1$ and $A_{10}$: either $A_1$ is ranked above $A_{10}$,or $A_{10}$ is ranked above $A_1$.Due to symmetry,the number of ways in which $A_1$ is above $A_{10}$ is equal to the number of ways in which $A_{10}$ is above $A_1$.Let $N$ be the number of ways where $A_1$ is above $A_{10}$. Then,the total number of ways is $N + N = 2N = 10!$.Therefore,$N = \frac{1}{2}(10!)$.

The number of ways in which ten candidates $A_1, A_2, ....... A_{10}$ can be ranked such that $A_1$ is always above $A_{10}$ is

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