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If (n^(n))/(n!)=(nx)/((n-1)!), x =...

If `(n^(n))/(n!)=(nx)/((n-1)!), x =`

A

`n^(n-2)`

B

`n^(n-1)`

C

`n^(n+1)`

D

`n^((1)/(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
`(n^(n))/(n!)=(n xx n^(n-1))/(n xx (n-1)!)=(n)/(n)xx(n^(n-1))/((n-1)!)`
`=1xx(n^(n-1))/((n-1)!)=(n^(n-1))/((n-1)!)`
So `nx = n^(n-1)`. Dividing both sides by n gets you `x = (n^(n-1))/(n)=n^(n-2)`. Remember that you must subtract exponents when dividing powers. Since the exponent of n is 1, you subtract that value from `n-1`.
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