The number of ordered pairs
Question:

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

1. (1) 3

2. (2) 2

3. (3) 6

4. (4) 4

Correct Option: , 4

Solution:

$\frac{36}{r+1} \times{ }^{35} \mathrm{C}_{r}\left(k^{2}-3\right)={ }^{35} \mathrm{C}_{r} \cdot 6$

$\Rightarrow \quad k^{2}-3=\frac{r+1}{6}$

$\Rightarrow \quad k^{2}=3+\frac{r+1}{6}$

$r$ can be 5,35 for $k \in I$

$r=5, k=\pm 2$

$r=35, k=\pm 3$

Hence, number of ordered pairs $=4$.