Aspire Faculty ID #16844 · Topic: AMU MCA 2024 · Just now
AMU MCA 2024

If $\frac{^nP_5}{^nP_3} = 20$, then the value of $n$ is:

Solution

$\frac{^nP_5}{^nP_3} = \frac{n!/(n-5)!}{n!/(n-3)!}$ $= \frac{(n-3)!}{(n-5)!}$ $= (n-3)(n-4)$ So, $(n-3)(n-4) = 20$ $n^2 - 7n + 12 = 20$ $n^2 - 7n - 8 = 0$ $(n-8)(n+1) = 0$ $n = 8$
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