Find n if { }^{n} P_{4}=18 cdot{ }^{n-1} P_{2 | vedclass

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(C) Given equation: ${ }^{n} P_{4}=18 \cdot{ }^{n-1} P_{2}$Using the formula ${ }^{n} P_{r} = \frac{n!}{(n-r)!}$,we have:$\frac{n!}{(n-4)!} = 18 \cdot

Vedclass · Accepted Answer

$6$

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(C) Given equation: ${ }^{n} P_{4}=18 \cdot{ }^{n-1} P_{2}$Using the formula ${ }^{n} P_{r} = \frac{n!}{(n-r)!}$,we have:$\frac{n!}{(n-4)!} = 18 \cdot \frac{(n-1)!}{(n-1-2)!}$$\frac{n(n-1)!}{(n-4)!} = 18 \cdot \frac{(n-1)!}{(n-3)!}$Dividing both sides by $(n-1)!$ and multiplying by $(n-4)!$:$n = 18 \cdot \frac{(n-4)!}{(n-3)!}$Since $(n-3)! = (n-3)(n-4)!$,we get:$n = \frac{18}{n-3}$$n(n-3) = 18$$n^2 - 3n - 18 = 0$$(n-6)(n+3) = 0$$n = 6$ or $n = -3$Since $n$ must be a positive integer and $n \ge 4$ for ${ }^{n} P_{4}$ to be defined,we discard $n = -3$.Therefore,$n = 6$.

Find $n$ if ${ }^{n} P_{4}=18 \cdot{ }^{n-1} P_{2}$.

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