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

Concept: A key identity of binomial coefficients is: \[ {^nC_{r+1}}=\frac{n-r}{r+1}{^nC_r} \] Also, \[ {^nC_r}=\frac{n}{n-r}{^{\,n-1}C_r} \] These relations help convert binomial coefficients into comparable forms.
Step 1:Express \(^{36}C_{r+1}\) in terms of \(^{35}C_r\).} First use \[ ^{36}C_{r+1}=\frac{36}{r+1}\,^{35}C_r \] Substitute into the given equation: \[ \frac{36}{r+1}\,^{35}C_r = \frac{6\,^{35}C_r}{k^2-3} \]
Step 2:Cancel the common term \(^{35}C_r\).} \[ \frac{36}{r+1}=\frac{6}{k^2-3} \] \[ 36(k^2-3)=6(r+1) \] \[ 6(k^2-3)=r+1 \] \[ r=6k^2-19 \]
Step 3:Use the bounds of binomial coefficients.} Since \[ ^{36}C_{r+1} \] exists only when \[ 0 \le r+1 \le 36 \] \[ 0 \le r \le 35 \] Substitute \(r=6k^2-19\): \[ 0 \le 6k^2-19 \le 35 \] \[ 19 \le 6k^2 \le 54 \] \[ \frac{19}{6} \le k^2 \le 9 \] Thus \[ k^2=4 \text{ or } 9 \] \[ k=\pm2,\pm3 \] Total integer values \(=4\). Hence the number of ordered pairs \((r,k)\) is \[ \boxed{4} \]