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(n+1)P/(n-1)P=72n=?

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home1h | Student, Undergraduate | (Level 2) eNoter

Posted June 6, 2012 at 4:55 PM via web

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(n+1)P/(n-1)P=72

n=?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 6, 2012 at 5:27 PM (Answer #1)

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You need to remember that the number of permutations of a set consisting of n elements is `n!,`  such that:

`(n+1)P = (n+1)!`

`(n-1)P = (n-1)!`

Hence, substituting `(n+1)!`  for `(n+1)P`  and `(n-1)!`  for `(n-1)P`  yields:

`((n+1)!)/((n-1)!) = 72`

You may write `(n+1)! = (n-1)!*n*(n+1)`  such that:

`((n-1)!*n*(n+1))/( (n-1)!) = 72`

Reducing like terms yields:

`n(n+1) = 72`

You need to open the brackets such that:

`n^2 + n = 72 =gt n^2 + n - 72 = 0`

You need to solve for natural n the equation  `n^2 + n - 72 = 0`  such that:

`n_(1,2) = (-1+-sqrt(1+288))/2`

`n_(1,2) = (-1+-sqrt289)/2`

`n_(1,2) = (-1+-17)/2`

`n_1 = (-1+17)/2 =gt n_1 = 8`

`n_2 = (-1-17)/2 =gt n_2 = -9`

Since the solution to equation needs to be a natural number, hence n = 8.

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