# If ${{\text{ }}^{2n}}{C_3}{{\text{:}}^n}{C_3} = 44:3$, find the value of n.

Hint: Expand the given combinations using the formula $^nC_r=\dfrac{n!}{r!(n-r)!}$ and simplify. Equate it with the RHS to find the value of n.

Complete step by step answer:

We have the given statement as

$ \dfrac{{^{2n}{C_3}}}{{^n{C_2}}} = \dfrac{{44}}{3}$

Now expanding these terms by using the combination formula $^nC_r=\dfrac{n!}{r!(n-r)!}$, we get

$ \Rightarrow 3 \times \dfrac{{2n!}}{{3!(2n - 3)!}} = \dfrac{{44n!}}{{2!(n - 2)!}} $

$ \Rightarrow \dfrac{3}{{3 \times 2}} \times \dfrac{{2n!}}{{(2n - 3)!}} = \dfrac{{44n!}}{{2 \times 1(n - 2)!}} $

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

$ \Rightarrow 2n(2n - 1)2(n - 1) = 44n(n - 1) $

$\Rightarrow n(n - 1)[4(2n - 1) - 44] = 0 $

$\Rightarrow n(n - 1)[8n - 4 - 44] = 0$

$\Rightarrow n(n - 1)[8n - 48] = 0 $

$\Rightarrow ({n^2} - n)(8n - 48) = 0 $

$\Rightarrow 8n(n - 1)(n - 6) = 0 $

$\Rightarrow n = 0,1,6 $

The value of n = 6 because n can neither be 0 nor be 1, as 1 would be less than the base value given in the question.

Therefore, n = 6 is the required solution.

Note: The large valued factorial terms must be removed if possible by suitable mathematical manipulations.