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PERMUTATIONS AND COMBINATIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSPERMUTATIONS AND COMBINATIONS2 Marks
Question
Determine n if $^{2n}{C_3}{\ : \ ^n}{C_3} = 11:1$
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Answer

Here,we have $^{2n}{C_3}{\ : \ ^n}{C_3} = 11:1$
$\Rightarrow \frac{2 \mathrm{n}_{\mathrm{C}_{3}}}{\mathrm{n}_{\mathrm{C}_{3}}}=\frac{11}{1}$
$ \Rightarrow \frac{{(2n)!}}{{3!(2n - 3)!}} \times \frac{{3!(n - 3)!}}{{n!}} = \frac{{11}}{1}$
$ \Rightarrow \frac{{(2n)(2n - 1)(2n - 2)(2n - 3)!}}{{3!(2n - 3)!}}$$ \times \frac{{3!(n - 3)!}}{{n(n - 1)(n - 2)(n - 3)!}} = \frac{{11}}{1}$
$\Rightarrow \frac{\frac{2 n \times(2 n-1) \times(2 n-2)}{3 !}}{\frac{n \times(n-1) \times(n-2)}{3 !}}=\frac{11}{1}$
$ \Rightarrow \frac{{(2n)(2n - 1)(2n - 2)}}{{n(n - 1)(n - 2)}} = \frac{{11}}{1}$
$\Rightarrow \frac{2 n \times(2 n-1) \times 2 \times(n-1)}{n \times(n-1) \times(n-2)}=\frac{11}{1}$
$\Rightarrow \frac{4 \times n \times(2 n-1)}{n \times(n-2)}=\frac{11}{1}$
$ \Rightarrow \frac{{4(2n - 1)}}{{n - 2}} = \frac{{11}}{1}$
$ \Rightarrow $ $4 \times(2 n-1)=11 \times(n-2)$
$ \Rightarrow 8n - 4 = 11n - 22$
$ \Rightarrow $ 3n = 18
$ \therefore$ n = 6

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