If {}^n{P_5} = 20 cdot {}^n{P_3},then n = | vedclass

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(B) Given the equation: ${}^n{P_5} = 20 \cdot {}^n{P_3}$Using the formula ${}^n{P_r} = \frac{n!}{(n-r)!}$,we have:$\frac{n!}{(n-5)!} = 20 \cdot \frac{

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$8$

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(B) Given the equation: ${}^n{P_5} = 20 \cdot {}^n{P_3}$Using the formula ${}^n{P_r} = \frac{n!}{(n-r)!}$,we have:$\frac{n!}{(n-5)!} = 20 \cdot \frac{n!}{(n-3)!}$Dividing both sides by $n!$ (since $n! 
eq 0$):$\frac{1}{(n-5)!} = \frac{20}{(n-3)!}$$(n-3)! = 20 \cdot (n-5)!$$(n-3)(n-4)(n-5)! = 20 \cdot (n-5)!$Since $(n-5)! 
eq 0$,we can cancel it from both sides:$(n-3)(n-4) = 20$$n^2 - 7n + 12 = 20$$n^2 - 7n - 8 = 0$$(n-8)(n+1) = 0$This gives $n = 8$ or $n = -1$.Since $n$ must be a positive integer and $n \ge 5$ for ${}^n{P_5}$ to be defined,we discard $n = -1$.Therefore,$n = 8$.

If ${}^n{P_5} = 20 \cdot {}^n{P_3}$,then $n = $

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