Key Concepts and Formulas

• Binomial Coefficient Definition: ${^nC_r} = \frac{n!}{r!(n-r)!}$
• Adjacent Binomial Coefficient Relation: ${^nC_{r+1}} = \frac{n}{r+1} {^{n-1}C_r}$
• Integer Constraints: Understanding when binomial coefficients are defined and the implications of k being an integer.

Step-by-Step Solution

Step 1: Understand the given equation and define the domain for $r$.

The given equation is: $6 \cdot {^{35}C_r} = (k^2 - 3) \cdot {^{36}C_{r+1}}$ For the binomial coefficients to be defined, we need $0 \le r \le 35$ for ${^{35}C_r}$ and $0 \le r+1 \le 36$ for ${^{36}C_{r+1}}$, which gives $-1 \le r \le 35$. Combining these, we have $0 \le r \le 35$. We are also given that $k$ is an integer.

Explanation: Defining the range of possible values for $r$ is crucial to ensure we only consider valid solutions.

Step 2: Apply the combination identity for simplification.

We use the identity ${^nC_{r+1}} = \frac{n}{r+1} \cdot {^{n-1}C_r}$ with $n=36$: ${^{36}C_{r+1}} = \frac{36}{r+1} \cdot {^{35}C_r}$ Substituting this into the original equation: $6 \cdot {^{35}C_r} = (k^2 - 3) \cdot \left( \frac{36}{r+1} \cdot {^{35}C_r} \right)$

Explanation: This substitution allows us to cancel out the ${^{35}C_r}$ term, simplifying the equation.

Step 3: Rearrange the equation to isolate $k^2-3$.

Since ${^{35}C_r} \neq 0$ for $0 \le r \le 35$, we can divide both sides by ${^{35}C_r}$: $6 = (k^2 - 3) \cdot \frac{36}{r+1}$ Multiply both sides by $(r+1)$: $6(r+1) = (k^2 - 3) \cdot 36$ Divide both sides by 36: $\frac{6(r+1)}{36} = k^2 - 3$ $\frac{r+1}{6} = k^2 - 3$

Explanation: Isolating $k^2 - 3$ allows us to establish a direct relationship between $r$ and $k$, which we can then use to find integer solutions for $k$.

Step 4: Analyze conditions for integer $k$.

Since $k$ is an integer, $k^2$ is a non-negative perfect square. Thus, $k^2 - 3$ must be an integer. Therefore, $\frac{r+1}{6}$ must be an integer. This implies $(r+1)$ must be a multiple of 6.

Also, since $k^2 \ge 0$, $k^2 - 3 \ge -3$. From the expression $\frac{r+1}{6}$, and knowing $r \ge 0$, we have $r+1 \ge 1$. So, $\frac{r+1}{6} \ge \frac{1}{6}$. Combining these, $k^2-3$ must be greater than or equal to -3. But since $\frac{r+1}{6}$ must be an integer, $k^2 - 3$ must be an integer.

Furthermore, $k^2 = \frac{r+1}{6} + 3$. Since $k^2$ must be a perfect square, $\frac{r+1}{6} + 3$ must be a perfect square.

Explanation: These conditions are essential for ensuring that $k$ is an integer.

Step 5: Find possible values of $r$.

From Step 1, we know $0 \le r \le 35$. This means $1 \le r+1 \le 36$. Since $r+1$ must be a multiple of 6 (from Step 4), the possible values for $r+1$ are: $6, 12, 18, 24, 30, 36$.

Let's find the corresponding values for $r$:

• If $r+1 = 6 \implies r=5$
• If $r+1 = 12 \implies r=11$
• If $r+1 = 18 \implies r=17$
• If $r+1 = 24 \implies r=23$
• If $r+1 = 30 \implies r=29$
• If $r+1 = 36 \implies r=35$

All these values of $r$ are within the valid domain $0 \le r \le 35$.

Explanation: We systematically find all possible values of $r$ that satisfy the divisibility condition.

Step 6: Determine corresponding $k$ values and count ordered pairs.

For each valid $r$, we calculate $k^2-3$ and then $k^2$. We only accept cases where $k^2$ is a perfect square.

1. For $r=5$: $k^2-3 = \frac{5+1}{6} = \frac{6}{6} = 1$ $k^2 = 1+3 = 4$ $k = \pm 2$. So, $(5, 2)$ and $(5, -2)$ are solutions.

2. For $r=11$: $k^2-3 = \frac{11+1}{6} = \frac{12}{6} = 2$ $k^2 = 2+3 = 5$ $k = \pm \sqrt{5}$. Not an integer, so no solution.

3. For $r=17$: $k^2-3 = \frac{17+1}{6} = \frac{18}{6} = 3$ $k^2 = 3+3 = 6$ $k = \pm \sqrt{6}$. Not an integer, so no solution.

4. For $r=23$: $k^2-3 = \frac{23+1}{6} = \frac{24}{6} = 4$ $k^2 = 4+3 = 7$ $k = \pm \sqrt{7}$. Not an integer, so no solution.

5. For $r=29$: $k^2-3 = \frac{29+1}{6} = \frac{30}{6} = 5$ $k^2 = 5+3 = 8$ $k = \pm \sqrt{8}$. Not an integer, so no solution.

6. For $r=35$: $k^2-3 = \frac{35+1}{6} = \frac{36}{6} = 6$ $k^2 = 6+3 = 9$ $k = \pm 3$. So, $(35, 3)$ and $(35, -3)$ are solutions.

Therefore, the ordered pairs $(r, k)$ are $(5, 2), (5, -2), (35, 3), (35, -3)$. There are 4 such ordered pairs.

Explanation: We explicitly check each possible value of $r$ to determine if it yields integer values for $k$.

Common Mistakes & Tips

• Remember to check for both positive and negative values of $k$ when taking the square root.
• Always ensure that the values of $r$ you obtain are within the valid range determined by the binomial coefficients.
• Don't forget the fundamental definition of binomial coefficients and when they are valid.

Summary

We simplified the given equation using the adjacent binomial coefficient relation. We then found the possible values of $r$ by enforcing the condition that $k$ must be an integer. Finally, we verified which of these values of $r$ actually resulted in integer values of $k$, counting the number of ordered pairs. The number of ordered pairs is 4.

Final Answer

The final answer is \boxed{4}, which corresponds to option (D).