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:

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Binomial Theorem

The number of ordered pairs

(r, k)

for which

6.^{35} C_{r} = (k^{2} - 3) \cdot^{36} C_{r + 1}

, where

k

is an integer, is:

Answer: (C)

Using

(r + 1 36) = \frac{36}{r + 1} (r 35)

, the equation

6 (r 35) = (k^{2} - 3) (r + 1 36)

reduces to

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

. For an integer

k

with

k^{2} > 3

k = \pm 2

gives

k^{2} - 3 = 1

, so

r = 5

; and

k = \pm 3

gives

k^{2} - 3 = 6

, so

r = 35

. Each value of

k^{2}

comes from two integer values of

k

, giving the ordered pairs

(5, 2), (5, - 2), (35, 3), (35, - 3)

— four in all.

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option