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

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

Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremProperties of Binomial coefficientsEasy2 minPYQ_2020

MathematicsEasysingle choice

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

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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 $= 4$

Stream:JEESubject:MathematicsTopic:Binomial TheoremSubtopic:Properties of Binomial coefficients

⏱ 2mℹ️ Source: PYQ_2020

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