6.Permutation and Combination
medium

$n$ का मान निकालिए, यदि

${ }^{2 n} C _{3}:{ }^{n} C _{3}=11: 1$

A

$6$

B

$6$

C

$6$

D

$6$

Solution

$\frac{^{2 n} C_{3}}{^{n} C_{3}}=\frac{11}{1}$

$\Rightarrow \frac{(2 n) !}{3 !(2 n-3) !} \times \frac{3 !(n-3) !}{n !}=11$

$\Rightarrow \frac{(2 n)(2 n-1)(2 n-2)(2 n-3) !}{(2 n-3) !} \times \frac{(n-3) !}{n(n-1)(n-2)(n-3) !}$

$\Rightarrow \frac{2(2 n-1)(2 n-2)}{(n-1)(n-2)}=11$

$\Rightarrow \frac{4(2 n-1)(n-1)}{(n-1)(n-2)}=11$

$\Rightarrow \frac{4(2 n-1)}{n-2}=11$

$\Rightarrow 4(2 n-1)=11(n-2)$

$\Rightarrow 8 n-4=11 n-22$

$\Rightarrow 11 n-8 n=-4+22$

$\Rightarrow 3 n=18$

$\Rightarrow n=6$

Standard 11
Mathematics

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