The number of ordered pairs (r, k) for which 6 . C _ r 35 = ( k ^ 2 - 3) . C _ r + 1 36 , where k is an integer is

Correct answer: D. 4 Given, 6 . C _ r 35 = ( k ^ 2 - 3) . C _ r + 1 36 ⇒ (36)/(r + 1) × ^ 35 C _ r ( k ^ 2 - 3) = 6 × ^ 35 C _ r , using properties of binomial coefficient. ⇒ k ^ 2 - 3 = (r + 1)/(6) ⇒ k ^ 2 = 3 + (r + 1)/(6) r can be 5,35 For r = 5, k = ± 2 r = 35, k = ± 3 Hence, number of solution = 4

JEE Main 20207 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 7 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2020 · 7 January · Evening Shift · Mathematics · Binomial Theorem

The number of ordered pairs (r, k) for which 6 . C _ r 35 = ( k ^ 2...

The number of ordered pairs

\left(r, k\right)

for which

6 . {C}_{r} 35 = \left({k}^{2} - 3\right) . {C}_{r + 1} 36 ,

where

k

is an integer is

Options

A $3$
B $2$
C $6$
D $4$

Correct answer

D. $4$

Step-by-step solution

Given, $6 . {C}_{r} 35 = \left({k}^{2} - 3\right) . {C}_{r + 1} 36$

$⇒ \dfrac{36}{r + 1}{\times}^{35}{C}_{r}\left({k}^{2} - 3\right) = 6{\times}^{35}{C}_{r}$ , using properties of binomial coefficient.
$⇒ {k}^{2} - 3 = \dfrac{r + 1}{6}$

$⇒ {k}^{2} = 3 + \dfrac{r + 1}{6}$

$r$ can be $5,35$

For $r = 5, k = \pm 2$

$r = 35, k = \pm 3$

Hence, number of solution $= 4$

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