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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.
Posted by sciencesolve on June 6, 2012 at 5:27 PM (Answer #1)
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