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PERMUTATIONS AND COMBINATIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSPERMUTATIONS AND COMBINATIONS2 Marks
Question
Determine n if $^{2n}{C_3}{:^n}{C_2} = 12:1$
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Answer

Here,we have $^{2 n} C_{3}:^{n} C_{3}=12: 1$
$\Rightarrow \frac{2 \mathrm{n}_{\mathrm{C}_{3}}}{\mathrm{n}_{\mathrm{C}_{3}}}=\frac{12}{1}$
$\Rightarrow \frac{\frac{2 n !}{3 !(2 n-3) !}}{\frac{n !}{3 !(n-3) !}}=\frac{12}{1}$
$\Rightarrow \frac{\frac{2 n \times(2 n-1) \times(2 n-2) \times(2 n-3) !}{3 !(2 n-3) !}}{\frac{n \times(n-1) \times(n-2) \times(n-3) !}{3 !(n-3) !}}=\frac{12}{1}$
$\Rightarrow \frac{\frac{2 n \times(2 n-1) \times(2 n-2)}{3 !}}{\frac{n \times(n-1) \times(n-2)}{3 !}}=\frac{12}{1}$
$\Rightarrow \frac{2 n \times(2 n-1) \times(2 n-2)}{n \times(n-1) \times(n-2)}=\frac{12}{1}$
$\Rightarrow \frac{2 n \times(2 n-1) \times 2 \times(n-1)}{n \times(n-1) \times(n-2)}=\frac{12}{1}$
$\Rightarrow \frac{4 \times n \times(2 n-1)}{n \times(n-2)}=\frac{12}{1}$
$\Rightarrow \frac{4 \times(2 n-1)}{(n-2)}=\frac{12}{1}$
$\Rightarrow 4 \times(2 n-1)=12 \times(n-2)$
$\Rightarrow$ 8 n - 4 = 12 n - 24
$\Rightarrow$ 12 n - 8 n = 24 - 4
$\Rightarrow$ 4 n = 20
$\therefore$ n = 5

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