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Permutations — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsPermutations3 Marks
Question
If P(5, r) = P(6, r - 1), find r
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Answer

We have,
P(5, r) = P(6, r - 1)
$\Rightarrow \frac{5!}{(5-\text{r})!}= \frac{6!}{\big[6-(\text{r}-1)\big]!}\ \Big[\because^\text{n}\text{p}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$
$\Rightarrow \frac{1}{(5-\text{r})!}=\frac{6}{[7-\text{r}]}!$
$\Rightarrow \frac{1}{(5-\text{r})!}= \frac{6}{(7-\text{r})\times(7-\text{r}-1)(7-\text{r}-2)!}$
$\Rightarrow \frac{1}{(5-\text{r})!}=\frac{6}{(7-\text{r})\times(6-\text{r})(5-\text{r}!)}$
$\Rightarrow 1 = \frac{6}{(7-\text{r})\times(6-\text{r})}$
$\Rightarrow(6-\text{r})\times(7-\text{r})= 6$
$\Rightarrow 42-6\text{r}-7\text{r}+\text{r}^2= 6$
$\Rightarrow \text{r}^2-12\text{r}+42-6=0$
$\Rightarrow \text{r}^2-9\text{r}-4\text{r}+36 = 0$
$\Rightarrow \text{r}(\text{r}-9)-4(\text{r}-9)= 0$
$\Rightarrow (\text{r}-9)(\text{r}-4)= 0$
$\Rightarrow \text{r}= 4 \begin{bmatrix}\because\text{r}\ \leq\ \text{n} \\ \therefore \ \text{r}-9\neq0 \end{bmatrix}$
Hence, $\text{r}= 4$

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