Find r, if 5 times ^{4}P_{r} = 6 times ^{5}P_ | vedclass

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(A) Given the equation: $5 	imes ^{4}P_{r} = 6 	imes ^{5}P_{r-1}$ Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have: $5 	imes \frac{4!}{(4-

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$r = 3$

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(A) Given the equation: $5 	imes ^{4}P_{r} = 6 	imes ^{5}P_{r-1}$ Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have: $5 	imes \frac{4!}{(4-r)!} = 6 	imes \frac{5!}{(5-(r-1))!} = 6 	imes \frac{5!}{(6-r)!}$ Since $5! = 5 	imes 4!$,we can write: $5 	imes \frac{4!}{(4-r)!} = 6 	imes \frac{5 	imes 4!}{(6-r)(5-r)(4-r)!}$ Dividing both sides by $5 	imes 4!$: $\frac{1}{(4-r)!} = \frac{6}{(6-r)(5-r)(4-r)!}$ $1 = \frac{6}{(6-r)(5-r)}$ $(6-r)(5-r) = 6$ $30 - 6r - 5r + r^{2} = 6$ $r^{2} - 11r + 24 = 0$ $(r-8)(r-3) = 0$ Thus,$r = 8$ or $r = 3$. However,for $^{4}P_{r}$,we must have $r \leq 4$. Therefore,$r = 8$ is rejected. Hence,$r = 3$.

Find $r,$ if $5 \times ^{4}P_{r} = 6 \times ^{5}P_{r-1}.$

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