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Question

Mathematics Question on permutations and combinations

Find the value of n such that nP5=42nP3,n>4^{n}P_{5} = 42\, ^{n}P_{3} , n > 4.

A

1010

B

1515

C

1212

D

2020

Answer

1010

Explanation

Solution

We have nP5=42nP3^{n}P_{5} = 42 ^{n}P_{3} n(n1)(n2)(n3)(n4)=42n(n1)(n2)\Rightarrow n\left(n - 1\right) \left(n - 2\right) \left(n - 3\right) \left(n - 4\right) = 42 n\left(n - 1\right) \left(n - 2\right) (n3)(n4)=42\Rightarrow \left(n-3\right)\left(n-4\right)= 42 [Since n>4n > 4, so n(n1)(n2)0n\left(n-1\right)\left(n-2\right) \ne0] n27n30=0\Rightarrow n^{2} - 7n - 30 = 0 n210n+3n30=0\Rightarrow n^{2} - 10n + 3n - 30 = 0 n=10 \Rightarrow n= 10 or n=3n= -3 As n cannot be negative, so n=10n = 10