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Sum

Show that (n + 1) `""^"n""P"_"r" = ("n" - "r" + 1) ""^(("n" + 1))"P"_"r"`

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#### Solution

L.H.S. = (n + 1) `""^"n""P"_"r" `

= `("n" + 1) xx ("n!")/(("n" - "r")!)`

= `(("n" + 1)!)/(("n" - "r")!)` .....(I)

= R.H.S. = `("n" - "r" + 1) ""^(("n" + 1))"P"_"r"`

= `("n" - "r" + 1) xx (("n" + 1)!)/(("n" - "r" + 1)!)`

= `(("n" - "r" + 1)("n" + 1)!)/(("n" - "r" + 1)("n" - "r")!)`

= `(("n" + 1)!)/(("n" - "r")!)` .....(II)

From (I) and (II), L.H.S. = R.H.S.

∴ (n + 1)`""^"n""P"_"r" = ("n" - "r" + 1) ""^(("n" + 1))"P"_"r"`

Concept: Permutations - Permutations When All Objects Are Distinct

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