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Permutations — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSPermutations5 Marks
Question
If P(15, r − 1) : P(16, r − 2) = 3 : 4, find r.
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Answer

We have,
P(15, r − 1) : P(16, r − 2) = 3 : 4
$\Rightarrow \frac{\text{p}(15,\text{r}-1)}{\text{p}(16,\text{r}-2)}=\frac{3}{4}$
$\Rightarrow\frac{\frac{15!}{[15-(\text{r}-1)]!}}{\frac{16!}{[16-(\text{r}-2)]!}}=\frac{3}{4}$
$\Rightarrow\frac{\frac{15!}{[16-\text{r}]!}}{\frac{16!}{[18-\text{r}]!}}=\frac{3}{4}$
$\Rightarrow\frac{15!}{(16-\text{r})!}\times\frac{(18-\text{r})!}{16!}=\frac{3}{4}$
$\Rightarrow \frac{15\times (18-\text{r})(17-\text{r})(16-\text{r})!}{(16-\text{r})!\times16\times15!}=\frac{3}{4}$
$\Rightarrow \frac{(18-\text{r})(17-\text{r})}{16}=\frac{3}{4}$
$\Rightarrow 306-18\text{r}-17\text{r}+\text{r}^2=\frac{3}{4}\times16$
$\Rightarrow \text{r}^2-35\text{r}+306=12$
$\Rightarrow \text{r}^2-35\text{r}+306-12=0$
$\Rightarrow \text{r}^2-35\text{r}+294=0$
$\Rightarrow \text{r}^2-21 \text{r}-14 \text{r}+294=0$
$\Rightarrow \text{r}( \text{r}-21)-14( \text{r}-21)=0$
$\Rightarrow ( \text{r}-21)( \text{r}-14)=0$
$\Rightarrow \text{r}-14=0$
$\Rightarrow \text{r}=14$
Hence, $ \text{r}=14$

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