The number of ordered pairs r,k for which 6.C_r35=k^2-3.C_(r+1)36, where k is an integer is

Correct answer: (D) 4. Given, 6.C_r35=k^2-3.C_(r+1)36 ⇒36/(r+1)×^35C_rk^2-3=6×^35C_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

Q48 / 75 · JEE Main 2020 (07 Jan Shift 2) · Mathematics

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

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

Solution

Given, $6 . {}^{35}{C_{r}} = (k^{2} - 3) . {}^{36}{C_{r + 1}}$

$\Rightarrow \frac{36}{r + 1} \times^{35} C_{r} (k^{2} - 3) = 6 \times^{35} C_{r}$ , using properties of binomial coefficient.
$\Rightarrow k^{2} - 3 = \frac{r + 1}{6}$

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

$r$ can be $5,35$

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

$r = 35, k = \pm 3$

Hence, number of solution $= 4$

Hence, the correct answer is option: 'D'.

The number of ordered pairs r,k for which 6.Cr35=k2-3.Cr+136, where k is an integer is

Solution

Relevant Questions

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