JEE PYQ: Permutation And Combination Question 25
Question 25 - 2020 (07 Jan Shift 2)
The number of ordered pairs $(r, k)$ for which $6 \cdot {}^{35}C_r = (k^2 - 3) \cdot {}^{36}C_{r+1}$, where $k$ is an integer, is:
(1) 3
(2) 2
(3) 6
(4) 4
Show Answer
Answer: (4)
Solution
$\frac{36}{r+1} \times {}^{35}C_r \cdot 6 \cdot \frac{r+1}{36} = (k^2-3){}^{35}C_r$. So $k^2 - 3 = \frac{r+1}{6}$, giving $k^2 = 3 + \frac{r+1}{6}$. For $k \in \mathbb{Z}$: $r = 5 \Rightarrow k = \pm 2$; $r = 35 \Rightarrow k = \pm 3$. Total ordered pairs $= 4$.
BETA
