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If nPr = 1814400 and nCr = 45, find n+4Cr+3 - Mathematics and Statistics

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Question

If nPr = 1814400 and nCr = 45, find n+4Cr+3 

Sum
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Solution

nPr = 1814400, nCr = 45

∴ `(""^"n""P"_"r")/(""^"n""C"_"r") = 1814400/45`

∴ `("n"!)/(("n" - "r")!) xx ("r"!("n" - "r")!)/("n"!) = 1814400/45`

∴ r! = 40320

∴ r! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

∴ r! = 8!

∴ r = 8

Also, nCr = 45

nC8 = 45

∴ `("n"!)/(8!("n" - 8)!)` = 45

∴ n(n – 1)(n – 2)(n – 3)(n – 4)(n – 5)(n – 6) (n – 7) = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3

Comparing on both sides, we get

n = 10

n+4Cr+3 = 14C11

= `(14!)/(11!(14 - 11)!)`

= `(14!)/(11!3!)`

= `(14 xx 13 xx 12 xx 11!)/(11! xx 3 xx 2 xx 1)`

= 364

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Properties of Combinations
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Chapter 3: Permutations and Combination - Exercise 3.6 [Page 65]

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