Question

The number of ordered pairs $(r,k)$ for which $6·{}^{35}C_r=\left(k^2-3\right){}^{36}C_{r+1}$, where $\mathrm{k}$ is an integer, is

• A. $4$
• B. $6$
• C. $2$
• D. $3$

Answer 1

The correct option is A

$4$

Explanation for the correct option:

Step 1: Simplify the given equation.

An equation $6·{}^{35}C_r=\left(k^2-3\right){}^{36}C_{r+1}$ is given.

Solve this given equation as follows:

$$\begin{gathered} 6·{}^{35}C_r=\left(k^2-3\right)\frac{36!}{(r+1)!(36-r-1)!} \\ \Rightarrow 6·{}^{35}C_r=\left(k^2-3\right)\frac{36!}{(r+1)!(35-r)!} \\ \Rightarrow 6·{}^{35}C_r=\left(k^2-3\right)\frac{36\left(35!\right)}{(r+1)\left[\left(r\right)!(35-r)!\right]} \\ \Rightarrow 6·{}^{35}C_r=\left(k^2-3\right)\frac{36}{(r+1)}{}^{35}C_r \\ \Rightarrow 6=\left(k^2-3\right)\frac{36}{(r+1)} \\ \Rightarrow \frac{r+1}{6}=\left(k^2-3\right) \\ \Rightarrow \frac{r+1}{6}+3=k^2 \end{gathered}$$

Step 2: Find the possible values of $r$.

Since, $k$ is an integer.

So, $r$ must be an integer.

Therefore, the possible values of $\frac{r+1}{6}=-6,-1,1,6$.

Since, $\frac{r+1}{6}+3=k^2$. So, $r$ can not be negative.

So, the values of $r$ are $5$ and $35$.

Since the number of values of $r$ is $2$.

Step 3: Find the possible values of $k$.

Since, $\frac{r+1}{6}+3=k^2$.

So, for $r=5$ find the value of $k$.

$$\begin{gathered} \frac{5+1}{6}+3=k^2 \\ \Rightarrow 1+3=k^2 \\ \Rightarrow 4=k^2 \\ \Rightarrow k=\pm 2 \end{gathered}$$

For $r=35$ find the value of $k$.

$$\begin{gathered} \frac{35+1}{6}+3=k^2 \\ \Rightarrow 6+3=k^2 \\ \Rightarrow 9=k^2 \\ \Rightarrow k=\pm 3 \end{gathered}$$

Therefore, the required ordered pairs are $(5,-2),(5,2),(35,-3)$ and $(35,3)$.

Therefore, The total number of required ordered pairs $(r,k)$ is $4$.

Hence, option (A) is the correct option.