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Show that (n + 1) (nPr) = (n – r + 1) [(n+1)Pr] - Mathematics and Statistics

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प्रश्न

Show that (n + 1) (nPr) = (n – r + 1) [(n+1)Pr]

योग
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उत्तर

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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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Permutations and Combination - Exercise 3.3 [पृष्ठ ५४]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 3 Permutations and Combination
Exercise 3.3 | Q 4 | पृष्ठ ५४
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