The number of elements in the set S = {(r, k) | vedclass

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(B) Given equation: $^{36}C_{r+1} = \frac{6(^{35}C_r)}{k^2-3}$.Using the property $^{n}C_r = \frac{n}{r} \cdot ^{n-1}C_{r-1}$, we have $^{36}C_{r+1} =

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$4$

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(B) Given equation: $^{36}C_{r+1} = \frac{6(^{35}C_r)}{k^2-3}$.Using the property $^{n}C_r = \frac{n}{r} \cdot ^{n-1}C_{r-1}$, we have $^{36}C_{r+1} = \frac{36}{r+1} \cdot ^{35}C_r$.Substituting this into the equation: $\frac{36}{r+1} \cdot ^{35}C_r = \frac{6(^{35}C_r)}{k^2-3}$.Assuming $^{35}C_r 
eq 0$, we get $\frac{36}{r+1} = \frac{6}{k^2-3}$, which simplifies to $\frac{6}{r+1} = \frac{1}{k^2-3}$.Thus, $k^2-3 = \frac{r+1}{6}$, or $k^2 = \frac{r+1}{6} + 3$.Since $k \in Z$, $k^2$ must be a perfect square integer. This implies $(r+1)$ must be a multiple of $6$.Given $0 \leq r \leq 35$, we have $1 \leq r+1 \leq 36$.Possible values for $r+1$ are $6, 12, 18, 24, 30, 36$.For $r+1 = 6$, $k^2 = 1+3 = 4 \implies k = \pm 2$. Pairs: $(5, 2), (5, -2)$.For $r+1 = 12$, $k^2 = 2+3 = 5$ (not a perfect square).For $r+1 = 18$, $k^2 = 3+3 = 6$ (not a perfect square).For $r+1 = 24$, $k^2 = 4+3 = 7$ (not a perfect square).For $r+1 = 30$, $k^2 = 5+3 = 8$ (not a perfect square).For $r+1 = 36$, $k^2 = 6+3 = 9 \implies k = \pm 3$. Pairs: $(35, 3), (35, -3)$.Total pairs $(r, k)$ are $(5, 2), (5, -2), (35, 3), (35, -3)$.There are $4$ such elements.

The number of elements in the set $S = \{(r, k) : k \in Z \text{ and } ^{36}C_{r+1} = \frac{6(^{35}C_r)}{(k^2-3)}\}$ is:

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