If 7 cdot ^nP_3 = 20 cdot ^{n+1}P_2,then n = | vedclass

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(D) Given the equation: $7 \cdot ^nP_3 = 20 \cdot ^{n+1}P_2$Using the formula $^nP_r = \frac{n!}{(n-r)!}$,we get:$7 \cdot \frac{n!}{(n-3)!} = 20 \cdot

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$6$

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(D) Given the equation: $7 \cdot ^nP_3 = 20 \cdot ^{n+1}P_2$Using the formula $^nP_r = \frac{n!}{(n-r)!}$,we get:$7 \cdot \frac{n!}{(n-3)!} = 20 \cdot \frac{(n+1)!}{(n-1)!}$$7 \cdot \frac{n!}{(n-3)!} = 20 \cdot \frac{(n+1)n!}{(n-1)(n-2)(n-3)!}$Dividing both sides by $n!$ and multiplying by $(n-3)!$:$7 = \frac{20(n+1)}{(n-1)(n-2)}$$7(n^2 - 3n + 2) = 20n + 20$$7n^2 - 21n + 14 = 20n + 20$$7n^2 - 41n - 6 = 0$Factoring the quadratic equation:$7n^2 - 42n + n - 6 = 0$$7n(n - 6) + 1(n - 6) = 0$$(7n + 1)(n - 6) = 0$Since $n \in \mathbb{N}$,we have $n = 6$.

If $7 \cdot ^nP_3 = 20 \cdot ^{n+1}P_2$,then $n = \dots$

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