If n, r are two positive integers such that 1 | vedclass

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(D) Given expression: ${ }^{n} P_{r+1} + (r+1) { }^{n-1} P_{r} + r^2 { }^{n-1} P_{r-1} + r { }^{n-1} P_{r-1}$ $= { }^{n} P_{r+1} + (r+1) { }^{n-1} P_{

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${ }^{n+1} P_{r+1}$

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(D) Given expression: ${ }^{n} P_{r+1} + (r+1) { }^{n-1} P_{r} + r^2 { }^{n-1} P_{r-1} + r { }^{n-1} P_{r-1}$ $= { }^{n} P_{r+1} + (r+1) { }^{n-1} P_{r} + (r^2+r) { }^{n-1} P_{r-1}$ $= \frac{n!}{(n-r-1)!} + (r+1) \frac{(n-1)!}{(n-r-1)!} + r(r+1) \frac{(n-1)!}{(n-r)!}$ $= \frac{n!(n-r) + (r+1)(n-1)!(n-r) + r(r+1)(n-1)!}{(n-r)!}$ $= \frac{(n-1)! [n(n-r) + (r+1)(n-r) + r(r+1)]}{(n-r)!}$ $= \frac{(n-1)! [n^2 - nr + nr - r^2 + n - r + r^2 + r]}{(n-r)!}$ $= \frac{(n-1)! [n^2 + n]}{(n-r)!} = \frac{n(n+1)(n-1)!}{(n-r)!} = \frac{(n+1)!}{(n-r)!} = { }^{n+1} P_{r+1}$

If $n, r$ are two positive integers such that $1 \leq r < n$,then ${ }^{n} P_{r+1} + r^2 { }^{n-1} P_{r-1} + (r+1) { }^{n-1} P_{r} + r { }^{n-1} P_{r-1} =$

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