Aspire Faculty ID #18344 · Topic: AMU MCA 2017 · Just now
AMU MCA 2017

If $^nP_4 = 20 \times ^nP_2$. Then, the value of $n$ is:

Solution

Given, $^nP_4 = 20 \times ^nP_2$$\frac{n!}{(n - 4)!} = 20 \times \frac{n!}{(n - 2)!}$$\frac{1}{(n - 4)!} = \frac{20}{(n - 2)(n - 3)(n - 4)!}$$(n - 2)(n - 3) = 20$$n^2 - 3n - 2n + 6 = 20$$n^2 - 5n + 6 - 20 = 0$$n^2 - 5n - 14 = 0$$(n - 7)(n + 2) = 0$This gives $n = 7$ or $n = -2$.Since $n$ must be a positive integer ($n \geq 4$ for $^nP_4$), $n = -2$ is not possible.$\therefore n = 7$
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