If ^nP_3 + ^nC_{n-2} = 14n,then n = | vedclass

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(A) Given the equation: $^nP_3 + ^nC_{n-2} = 14n$ We know that $^nP_3 = \frac{n!}{(n-3)!} = n(n-1)(n-2)$ And $^nC_{n-2} = ^nC_2 = \frac{n(n-1)}{2}$ Su

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$5$

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(A) Given the equation: $^nP_3 + ^nC_{n-2} = 14n$ We know that $^nP_3 = \frac{n!}{(n-3)!} = n(n-1)(n-2)$ And $^nC_{n-2} = ^nC_2 = \frac{n(n-1)}{2}$ Substituting these into the equation: $n(n-1)(n-2) + \frac{n(n-1)}{2} = 14n$ Since $n \geq 3$,we can divide by $n$: $(n-1)(n-2) + \frac{n-1}{2} = 14$ Multiply by $2$: $2(n^2 - 3n + 2) + n - 1 = 28$ $2n^2 - 6n + 4 + n - 1 = 28$ $2n^2 - 5n - 25 = 0$ Factoring the quadratic: $(2n + 5)(n - 5) = 0$ Since $n$ must be a positive integer,$n = 5$.

If $^nP_3 + ^nC_{n-2} = 14n$,then $n = $

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