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 sp

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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 specifically $^{n+1} C_{r+1} = \frac{n+1}{r+1} \cdot ^{n} C_{r}$,we can rewrite the equation:$k^2 - 3 = \frac{6 \cdot ^{35} C_{r}}{^{36} C_{r+1}} = \frac{6 \cdot ^{35} C_{r}}{\frac{36}{r+1} \cdot ^{35} C_{r}} = \frac{6(r+1)}{36} = \frac{r+1}{6}$.Since $k$ is an integer,$k^2 - 3$ must be a non-negative integer. Also,for the combinations to be defined,$0 \le r \le 35$.Thus,$k^2 = \frac{r+1}{6} + 3 = \frac{r+19}{6}$.For $k^2$ to be a perfect square,we test values of $r \in \{0, 1, ..., 35\}$:If $r=5$,$k^2 = \frac{5+19}{6} = 4 \Rightarrow k = \pm 2$.If $r=35$,$k^2 = \frac{35+19}{6} = 9 \Rightarrow k = \pm 3$.The ordered pairs $(r, k)$ are $(5, 2), (5, -2), (35, 3), (35, -3)$.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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