10 persons stand in a line which include Ram and Shyam. Let x be the number of ways in which Ram is ahead of Shyam and y be the number of ways in which Shyam is ahead of Ram. Then

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10 persons stand in a line which include Ram and Shyam. Let x be the number of ways in which Ram is ahead of Shyam and y be the number of ways in which Shyam is ahead of Ram. Then

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$x + y = 10!$

$x \neq y$

$x = \frac{10!}{2!}$

$x = 9!$

answer is A, C.

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Detailed Solution

x= No of ways in which Ram is ahead of Shyam= $\frac{n!}{r!} = \frac{10!}{2!}$

y= No of ways in which Shyam is ahead of Ram= $\frac{n!}{r!} = \frac{10!}{2!}$

x+y= $\frac{10!}{2!} + \frac{10!}{2!} = 10!$ ,x=y

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