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Question

Find n if nP4=18n1P2
  1. 6
  2. 12
  3. 4
  4. 8


Solution

The correct option is A 6
nP4=18n1P2
\Rightarrow \frac{n!}{(n-4)!} = 18. \frac{(n-1)!}{(n-1-2)!}\\
\Rightarrow \frac{n!}{(n-4)!} = 18. \frac{(n-1)!}{(n-3)!}\\
\Rightarrow \frac{n(n-1)!}{(n-4)!} = 18. \frac{(n-1)!}{(n-3)(n-4)!}\\
\therefore n = \frac{18}{n-3}\\
i.e.,~n^2 - 3n - 18 = 0 \Rightarrow (n-6)(n+3)=0\\
\Rightarrow n = 6, -3\)
But n cannot be negative
n=6

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