Question

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

• A. \(2\)
• B. \(4\)
• C. \(8\)
• D. \(16\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We use the binomial coefficient identity \({}^{n}C_{r} = \frac{n}{r} \cdot {}^{n-1}C_{r-1}\) to simplify the equation.
Then, we analyze the resulting integer constraints on \(k\) and the range constraints on \(r\).
Step 2: Key Formula or Approach:
1. \({}^{36}C_{r+1} = \frac{36}{r+1} \cdot {}^{35}C_r\).
2. Range of \(r\): \(0 \leq r \leq 35\) for \({}^{35}C_r\) to be defined, and \(0 \leq r+1 \leq 36\) for \({}^{36}C_{r+1}\).
Step 3: Detailed Explanation:
Substitute the identity into the given equation:
\[ \frac{36}{r+1} \cdot {}^{35}C_r = \frac{6}{k^2-3} \cdot {}^{35}C_r \]
Since binomial coefficients are non-zero in the defined range, we can divide both sides:
\[ \frac{36}{r+1} = \frac{6}{k^2-3} \]
\[ \frac{6}{r+1} = \frac{1}{k^2-3} \Rightarrow k^2 - 3 = \frac{r+1}{6} \]
\[ k^2 = 3 + \frac{r+1}{6} \]
Since \(k \in \mathbb{Z}\), \(k^2\) must be a perfect square.
Also, \(0 \leq r \leq 35 \Rightarrow 1 \leq r+1 \leq 36\). Thus \(0<\frac{r+1}{6} \leq 6\).
Possible values for \(\frac{r+1}{6}\):
- If \(\frac{r+1}{6} = 1 \Rightarrow r=5\). Then \(k^2 = 3+1 = 4 \Rightarrow k = \pm 2\).
Solutions: \((5, 2), (5, -2)\).
- If \(\frac{r+1}{6} = 2, 3, 4, 5\), then \(k^2 = 5, 6, 7, 8\). No integer \(k\) exists.
- If \(\frac{r+1}{6} = 6 \Rightarrow r=35\). Then \(k^2 = 3+6 = 9 \Rightarrow k = \pm 3\).
Solutions: \((35, 3), (35, -3)\).
Total solutions in set S = 4.
Step 4: Final Answer:
The number of elements in set S is 4.