(D) Total number of ways to arrange $10$ candidates is $10!$.In any arrangement,there are only two possibilities for the relative positions of $A_1$ a

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(D) Total number of ways to arrange $10$ candidates is $10!$.In any arrangement,there are only two possibilities for the relative positions of $A_1$ and $A_{10}$: either $A_1$ is above $A_{10}$ or $A_{10}$ is above $A_1$.Since these two cases are symmetric,the number of ways where $A_1$ is above $A_{10}$ is exactly half of the total number of arrangements.Therefore,the required number of ways is $\frac{10!}{2} = \frac{1}{2}(10!)$.

Class 11 Mathematics Permutation and Combination Definition of permutation, Number of permutations with or without repetition, Conditional permutations

Medium

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

A
$5!$
B
$2(5!)$
C
$10!$
D
$\frac{1}{2}(10!)$

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Permutation and Combination

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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

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