If ${}^{2n}{C}_3:{}^n{C}_2=44:3$, then for which of the following values of $r$, the value of ${}^n{C}_r$ will be $15$?

The correct option is B

$r=4$

Explanation of the correct answer:

Step1:Find value of $n$

A combination of $r$ things from $n$ things can be given as ${}^n{C}_r$ such that

${}^n{C}_r=\frac{n!}{(n-r)!r!}$.

It is given that

${}^{2n}{C}_3:{}^n{C}_2=44:3$

It can be written as

$\frac{{}^{2n}{C}_3}{{}^{n}C_2}=\frac{44}{3}$ …..(i)

$\Rightarrow$$\frac{{\displaystyle \frac{\left(2n\right)!}{(2n-3)!3!}}}{{\displaystyle \frac{n!}{(n-2)!2!}}}=\frac{44}{3}$

By expanding the factorials, it can be written as

$\frac{{\displaystyle \frac{\left(2n\right)(2n-1)(2n-2)}{3!}}}{{\displaystyle \frac{n(n-1)}{2!}}}=\frac{44}{3}$

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

$\Rightarrow$$\left(2n\right)(2n-1)(2n-2)=44n(n-1)$ …...(ii)

Equation (ii) can be simplified as

$(2n-1)(2n-2)=22(n-1)$

$\Rightarrow$$2(2n-1)(n-1)=22(n-1)$

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

$\Rightarrow$$n=6$

Step 2: Using the value of $n$ to find the value of $r$

According to the question,

${}^n{C}_r=15$

${}^6{C}_r=15$ …..(iii)

Also, it is known that

${}^{6}C_2=15={}^{6}C_4$

So,${}^6{C}_r={}^6{C}_2={}^{6}C_4$

$\mathbf{\therefore }\mathit{r}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{4}$

Also $2$ is not in the options given.

Hence, Option (B) is correct.