If 5 P(4, n) = 6.P(5, r - 1), find n. - Vidyadip

Question

If 5 P(4, n) = 6.P(5, r - 1), find n.

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Answer

We have, 5 P(4, n) = 6.P(5, r - 1) $\Rightarrow 5\times\frac{4!}{(4-\text{n})!}=6\times \frac{5!}{\big[5-(\text{r}-1)\big]!}\ \Big[\because^\text{n}\text{p}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$ $\Rightarrow5\times \frac{4}{(4-\text{n})!}=\frac{6\times5\times4!}{[5-\text{n}+1]}!$ $\Rightarrow \frac{1}{(4-\text{n})!}= \frac{6}{[6-\text{n}]!}$ $\Rightarrow \frac{1}{(4-\text{n})!}=\frac{6}{(6-\text{n})\times(6-\text{n}-1)(6-\text{n}-2)!}$ $\Rightarrow \frac{1}{(4-\text{n})}= \frac{6}{(6-\text{n})(5-\text{n})(4-\text{n})!}$ $\Rightarrow \frac{(6-\text{n})(5-\text{n})(4-\text{n})!}{(4-\text{n})!}=6$ $\Rightarrow(6-\text{n})\times(5-\text{n})= 6$ $\Rightarrow 30-6\text{n}-5\text{n}+\text{n}^2= 6$ $\Rightarrow \text{n}^2-11\text{n}+30=6$ $\Rightarrow \text{r}^2-11\text{r}+24 = 0$ $\Rightarrow \text{n}^2-8\text{n}-3\text{n}+24=0 $ $\Rightarrow \text{n}(\text{n}-8)-3(\text{n}-8)= 0$ $\Rightarrow (\text{n}-8)(\text{n}-3)= 0$ $\Rightarrow \text{n}-3 = 0$ $\Rightarrow \text{r}= 3 \begin{bmatrix}\because\text{n}\ \leq\ 4 \\ \therefore \ \text{n}\neq8 \end{bmatrix}$ Hence, $\text{r}= 3$

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