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Permutations — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsPermutations5 Marks
Question
If $^\text{n+5}\text{P}_\text{n+1}=-\frac{11(\text{n-1})}{2}\ ^\text{n+3}\text{P}_\text{n}$ find n.
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Answer

We have,
$^\text{n+5}\text{P}_\text{n+1}=-\frac{11(\text{n-1})}{2}\ ^\text{n+3}\text{P}_\text{n}$
$\Rightarrow \frac{​​​​​​​(​\text{n+5)}!}{[\text{n+5}-(\text{n+1})]!}=\frac{11(\text{n-1})}{2}\times\frac{(\text{n}+3)!}{[\text{n}+3-\text{n}]!}$
$\Rightarrow \frac{​​​​​​​(​\text{n+5)}!}{[\text{n+5}-\text{n-1}]!}=\frac{11(\text{n-1})}{2}\times\frac{(\text{n}+3)}{3!}$
$\Rightarrow \frac{​​​​​​​(​\text{n+5)}!}{4!}=\frac{11(\text{n-1})}{2}\times\frac{(\text{n}+3)!}{3!}$
$\Rightarrow \frac{​​​​​​​(​\text{n+5)}(​\text{n+4)}(​\text{n+3)}!}{4!}=\frac{11(\text{n-1})}{2}\times\frac{(\text{n}+3)!}{3!}$
$\Rightarrow \frac{​​​​​​​(​\text{n+5)}(​\text{n+4)}!}{4\times3!}=\frac{11(\text{n-1})}{2\times3}$
$\Rightarrow (\text{n}+5)(\text{n}+4)=\frac{11(\text{n}-1)\times4}{2}$
$\Rightarrow (\text{n}+5)(\text{n}+4)=22(\text{n}-1)$
$\Rightarrow \text{n}^2+4\text{n}+5\text{n}+20=22\text{n}-22$
$\Rightarrow \text{n}^2+9\text{n}-22\text{n}+20+22=0$
$\Rightarrow \text{n}^2-13\text{n}+42=0$
$\Rightarrow \text{n}^2-6\text{n}-7\text{n}+42=0$
$\Rightarrow \text{n}(\text{n}-6)-7(\text{n}-6)=0$
$\Rightarrow \text{n}=6 \ \text{or},\ \text{n}=7$
Hence, $\text{n} =6\ \text{or},\ 7$

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