Question

If $^{36}C_{r+1} = 6 \cdot \frac{^{35}C_r}{k^2 - 3}$, then number of ordered pairs $(r, k)$, where $r, k \in Z$ is ______.

Answer 1

Step 1: Understanding the Concept:
We use the binomial property \( nC_r = \frac{n}{r} (n-1)C_{r-1} \) to simplify the equation.
Step 2: Key Formula or Approach:
\( 36C_{r+1} = \frac{36}{r+1} \cdot 35C_r \).
Step 3: Detailed Explanation:
Substituting this into the given equation:
\[ \frac{36}{r+1} \cdot 35C_r = \frac{6 \cdot 35C_r}{k^2 - 3} \]
Assuming \( 35C_r \neq 0 \) (which is true for valid \( r \)):
\[ \frac{6}{r+1} = \frac{1}{k^2 - 3} \implies k^2 - 3 = \frac{r+1}{6} \implies k^2 = 3 + \frac{r+1}{6} \]
Since \( k \) and \( r \) are integers, \( r+1 \) must be a multiple of 6.
Also, \( 0 \le r \le 35 \), so \( 1 \le r+1 \le 36 \).
Possible values of \( r+1 \in \{6, 12, 18, 24, 30, 36\} \).
Check for perfect square \( k^2 \):
- If \( r+1 = 6 \), \( k^2 = 3 + 1 = 4 \implies k = \pm 2 \). (2 pairs: \( (5, 2), (5, -2) \))
- If \( r+1 = 36 \), \( k^2 = 3 + 6 = 9 \implies k = \pm 3 \). (2 pairs: \( (35, 3), (35, -3) \))
Other values do not result in perfect squares.
Step 4: Final Answer:
Total number of ordered pairs is 4.