Question
The number of elements in the set $S = \left\{ (r,k):k \in \mathbf{Z} \right.\ $ and $\left. \ \ ^{36}C_{r + 1} = \frac{6\left( \ ^{35}C_{r} \right)}{\left( k^{2} - 3 \right)} \right\}$, is
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Solution
Sol. [Key Concept: Reduce the combination ratio to $k^{2} = (r + 19)/6$; count integer pairs $(r,k)$ with $0 \leq \ r \leq \ 35$.]
$$\frac{36}{r + 1} \cdot \ ^{35}C_{r} = 6 \cdot \frac{\ ^{35}C_{r}}{k^{2} - 3}$$
$$\begin{matrix} & k^{2} - 3 = \frac{r + 1}{6} \\ & k^{2} = \frac{r + 19}{6} \\ & 0 \leq r \leq 35 \\ & 19 \leq r + 19 \leq 54 \\ & 3.1 \leq \frac{r + 19}{6} \leq 9 \\ & \ \therefore k^{2} = 4,9 \\ & k = \pm 2, \pm 3 \\ & \ \therefore(r,k) \Rightarrow (5, \pm 2) \Rightarrow (35, \pm 3) \\ & \ \therefore 4 \end{matrix}$$
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