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Permutations — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSPermutations4 Marks
Question
If P(2n − 1, n) : P(2n + 1, n − 1) = 22 : 7 find n.
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Answer

We have, P(2n − 1, n) : P(2n + 1, n − 1) = 22 : 7 $\Rightarrow \frac{\text{p}(\text{2n-1,n})}{\text{P} (\text{2n}+1,\text{n} -1)}=\frac{22}{7}$ $\Rightarrow\frac{\frac{\text{(2n-1)!}}{(\text{(2n-1-n)!})}}{\frac{\text{(2n+1)}!}{\text{[2n+1-(n-1)]!}}}=\frac{22}{7}$ $\Rightarrow \frac{\text{(2n-1) }\times\text{(n+2)!}}{\text{(n-1)!}(\text{2n+1})!}=\frac{22}{7}$ $\Rightarrow \frac{\text{(2n-1)}\times\text{(n+2)(n+2-1(n+2-2)(n+2-3)!}}{\text{(n+1)!(2n+1).2n.(2n+1)!}}=\frac{22}{7}$ $\Rightarrow \frac{\text{n(n+2)(n+1)}}{2\text {(2n+1)}}=\frac{22}{7}$ $\Rightarrow \frac{\text{n}^2+\text{n}+\text{2n}+2}{4\text{n}+2}=\frac{22}{7}$ $\Rightarrow 7(\text{n}^2+3\text{m}+2)=22\times (4\text{n}+2)$ $\Rightarrow7\text{n}^2+21\text{n}+44=88\text{n}-44$ $\Rightarrow7\text{n}^2+21\text{n}-88\text{n}+14-44=0$ $\Rightarrow 7\text{n}^2-67\text{n}-30=0$ $\Rightarrow 7\text{n}^2 -70\text{n}+3\text{n}-30=0$ $\Rightarrow 7\text{n}(\text{n}-10)+3(\text{n}-10)=0$ $\Rightarrow (\text{n}-10)(7\text{n}+3)=0$ $\Rightarrow \text{n}-10=0$ $\Rightarrow \text{n}=10\ [\because \text{7n+3}\neq 0]$

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