The number of ordered pairs (r, k) for which | vedclass

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(C) Given the equation: $6 \cdot ^{35}C_{r} = (k^{2} - 3) \cdot ^{36}C_{r+1}$.Using the identity $^{n}C_{r} = \frac{n}{r} \cdot ^{n-1}C_{r-1}$ or $^{n

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$4$

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(C) Given the equation: $6 \cdot ^{35}C_{r} = (k^{2} - 3) \cdot ^{36}C_{r+1}$.Using the identity $^{n}C_{r} = \frac{n}{r} \cdot ^{n-1}C_{r-1}$ or $^{n+1}C_{r+1} = \frac{n+1}{r+1} \cdot ^{n}C_{r}$,we have $^{36}C_{r+1} = \frac{36}{r+1} \cdot ^{35}C_{r}$.Substituting this into the equation:$6 \cdot ^{35}C_{r} = (k^{2} - 3) \cdot \frac{36}{r+1} \cdot ^{35}C_{r}$.Since $^{35}C_{r} 
eq 0$,we can divide both sides by $^{35}C_{r}$:$6 = (k^{2} - 3) \cdot \frac{36}{r+1}$.Rearranging for $(k^{2} - 3)$:$k^{2} - 3 = \frac{6(r+1)}{36} = \frac{r+1}{6}$.Thus,$k^{2} = \frac{r+1}{6} + 3 = \frac{r+19}{6}$.Since $k$ is an integer,$k^{2}$ must be a perfect square. Also,$0 \le r \le 35$.For $r=5$,$k^{2} = \frac{5+19}{6} = 4 \Rightarrow k = \pm 2$.For $r=35$,$k^{2} = \frac{35+19}{6} = 9 \Rightarrow k = \pm 3$.These give the ordered pairs $(5, 2), (5, -2), (35, 3), (35, -3)$.There are $4$ such ordered pairs.

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

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