Question

${}^n{C}_{r-1}=36,{}^n{C}_r=84\&{}^n{C}_{r+1}=126$, then $r$ is equal to:

• A. $1$
• B. $2$
• C. $3$
• D. None of these

Answer 1

The correct option is C

$3$

Explanation for the correct option:

Finding the value of $r$:

Given that:

${}^n{C}_{r-1}=36,{}^n{C}_r=84\&{}^n{C}_{r+1}=126$

Now, $\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{84}{36}....\left(i\right)$

We know that ${}^n{C}_r=\frac{n!}{r!(n-r)!}$

$\therefore {}^n{C}_{r-1}=\frac{n!}{(r-1)!(n-r+1)!}$

Applying and solving $\left(i\right)$, we get;

$$\begin{gathered} \Rightarrow \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}}=\frac{84}{36} \\ \Rightarrow \frac{\frac{n!}{r(r-1)!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)(n-r)!}}=\frac{7}{3} \\ \Rightarrow \frac{(n-r+1)}{r}=\frac{7}{3} \\ \Rightarrow 3n-3r+3=7r \\ \Rightarrow 10r=3n+3....\left(ii\right) \end{gathered}$$

Now, $\frac{{}^n{C}_{r+1}}{{}^n{C}_r}=\frac{126}{84}$

$$\begin{gathered} \Rightarrow \frac{\frac{n!}{\left(r+1\right)!(n-r-1)!}}{\frac{n!}{r!(n-r)!}}=\frac{126}{84} \\ \Rightarrow \frac{\frac{n!}{\left(r+1\right)r!(n-r-1)!}}{\frac{n!}{r!(n-r)(n-r-1)!}}=\frac{3}{2} \\ \Rightarrow \frac{(n-r)}{(r+1)}=\frac{3}{2} \\ \Rightarrow 5r=2n-3.....\left(iii\right) \end{gathered}$$

Now, dividing equation $\left(ii\right)$ by $\left(iii\right)$,

$$\begin{gathered} \Rightarrow \frac{10r}{5r}=\frac{(3n+3)}{(2n-3)} \\ \Rightarrow 2(2n-3)=3n+3 \\ \Rightarrow 4n-6=3n+3 \\ \Rightarrow n=9 \\ \therefore 10r=3n+3 \\ =3\left(9\right)+3 \\ \Rightarrow 10r=30 \\ \Rightarrow r=3 \end{gathered}$$

Hence, the correct answer is option (C).