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Question
Show that (n + 1) (nPr) = (n – r + 1) [(n+1)Pr]
Sum
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Solution
L.H.S. = (n + 1) (nPr)
= `("n"+ 1)("n"!)/(("n" - "r")!)`
= `(("n" + 1)!)/(("n" - "r")!)` ...(1)
R.H.S. = (n – r + 1) [(n+1)Pr]
= `("n" - "r" + 1)* (("n" + 1)!)/(("n" + 1 - "r")!)`
= `("n" - "r" + 1)* (("n" + 1)!)/(("n" - "r" + 1)!)`
= `("n" - "r" + 1)* (("n" + 1)!)/(("n" - "r" + 1)("n" - "r")!)`
= `(("n" + 1)!)/(("n" - "r")!)` ...(2)
From (1) and (2), we get, L.H.S. = R.H.S.
Hence, (n + 1) (nPr) = (n – r + 1) [(n+1)Pr]
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