Question

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Answer 1

GIVEN :–

$\\ \bf \implies ^{2n}C_{3}: \: ^{n}C_{2} = 44:3$

TO FIND :–

• Value of 'n' = ?

SOLUTION :–

$\\ \bf \implies ^{2n}C_{3}: \: ^{n}C_{2} = 44:3$

$\\ \bf \implies \dfrac{^{2n}C_{3}}{^{n}C_{2}} = \dfrac{44}{3}$

• We know that –

$\\ \large\implies \red{ \boxed{ \bf ^{n}C_{r} = \dfrac{n!}{(n - r)!(r)!}}}$

• So that –

$\\ \bf \implies \dfrac{\dfrac{(2n)!}{(2n -3)!(3)!}}{\dfrac{n!}{(n - 2)!(2)!}} = \dfrac{44}{3}$

$\\ \bf \implies \dfrac{\dfrac{(2n)(2n - 1)(2n - 2)(2n - 3)!}{(2n -3)!(3)!}}{\dfrac{n(n - 1)(n - 2)!}{(n - 2)!(2)!}} = \dfrac{44}{3}$

$\\ \bf \implies \dfrac{\dfrac{(2n)(2n - 1)(2n - 2)}{(3)(2)(1)}}{\dfrac{n(n - 1)}{(2)(1)}} = \dfrac{44}{3}$

$\\ \bf \implies\dfrac{2(2n)(2n - 1)(2n - 2)}{6n(n - 1)}= \dfrac{44}{3}$

$\\ \bf \implies\dfrac{2(2n)(2n - 1)(n - 1)}{3n(n - 1)}= \dfrac{44}{3}$

$\\ \bf \implies\dfrac{2(2n)(2n - 1)}{3n}= \dfrac{44}{3}$

$\\ \bf \implies\dfrac{2(2n)(2n - 1)}{n}=44$

$\\ \bf \implies4(2n - 1)=44$

$\\ \bf \implies(2n - 1)= \dfrac{44}{4}$

$\\ \bf \implies2n -1=11$

$\\ \bf \implies2n=11 + 1$

$\\ \bf \implies2n=12$

$\\ \bf \implies n= \dfrac{12}{2}$

$\\ \large\implies \red{ \boxed{ \bf n=6}}$

• Hence , Option (1) is correct.