Find r if ^{5}P_{r} = 2 times ^{6}P_{r-1}. | vedclass

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(A) Given equation: $^{5}P_{r} = 2 	imes ^{6}P_{r-1}$Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:$\frac{5!}{(5-r)!} = 2 	imes \frac{6!

Vedclass · Accepted Answer

$r = 3$

Vedclass · Answer

(A) Given equation: $^{5}P_{r} = 2 	imes ^{6}P_{r-1}$Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:$\frac{5!}{(5-r)!} = 2 	imes \frac{6!}{(6-(r-1))!}$$\frac{5!}{(5-r)!} = \frac{2 	imes 6!}{(7-r)!}$$\frac{5!}{(5-r)!} = \frac{2 	imes 6 	imes 5!}{(7-r)(6-r)(5-r)!}$Dividing both sides by $5!$ and multiplying by $(5-r)!$:$1 = \frac{12}{(7-r)(6-r)}$$(7-r)(6-r) = 12$$42 - 7r - 6r + r^{2} = 12$$r^{2} - 13r + 30 = 0$$(r-10)(r-3) = 0$So,$r = 10$ or $r = 3$.Since the condition for $^{n}P_{r}$ is $0 \leq r \leq n$,for $^{5}P_{r}$,we must have $r \leq 5$.Therefore,$r = 3$ is the only valid solution.

Find $r$ if $^{5}P_{r} = 2 \times ^{6}P_{r-1}$.

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