Question

All the solutions \((n,r)\) of the equation \( \frac{{}^nC_r}{{}^{n+1}C_r} = \frac{1}{3} \) can be obtained from one of the following equations given in the options for \( k=1,2,3, \dots \):

• A. \( \frac{{}^{3k}C_{2k}}{{}^{3k+1}C_{2k}} = \frac{1}{3} \)
• B. \( \frac{{}^{3k-1}C_{2k}}{{}^{3k}C_{2k}} = \frac{1}{3} \)
• C. \( \frac{{}^{4k}C_{2k}}{{}^{4k+1}C_{2k}} = \frac{1}{3} \)
• D. \( \frac{{}^{4k-1}C_{2k}}{{}^{4k}C_{2k}} = \frac{1}{3} \)

Hint:

Whenever dealing with the ratio of binomial coefficients, immediately simplify using the \( \frac{n-r+1}{n+1} \) identity to turn the combinatorial problem into a simple algebraic one.

Answer 1

The Correct Option is A

Step 1: Simplify the combination ratio.
Use $\dfrac{{}^nC_r}{{}^{n+1}C_r}$. Writing factorials, $\dfrac{n!}{r!(n-r)!}\cdot\dfrac{r!(n+1-r)!}{(n+1)!} = \dfrac{n+1-r}{n+1}$.
Step 2: Set the ratio equal to $\tfrac13$.
\[ \frac{n+1-r}{n+1} = \frac{1}{3}. \]
Step 3: Clear the fraction.
Cross-multiply: $3(n+1-r) = n+1$, so $3n + 3 - 3r = n + 1$, giving $2n + 2 = 3r$, i.e. $2(n+1) = 3r$.
Step 4: Force $r$ to be even.
Since $2(n+1) = 3r$, the right side must be even, so $r$ is even. Write $r = 2k$.
Step 5: Solve for $n$.
Then $2(n+1) = 3(2k) = 6k$, so $n + 1 = 3k$, giving $n = 3k - 1$. So solutions are $(n,r) = (3k-1,\ 2k)$.
Step 6: Match the option form.
With $n = 3k-1$ and $r = 2k$, the family is generated by $\dfrac{{}^{3k}C_{2k}}{{}^{3k+1}C_{2k}} = \dfrac13$ (the same relation indexed so $n+1 = 3k$), which is option (A).
\[ \boxed{\dfrac{{}^{3k}C_{2k}}{{}^{3k+1}C_{2k}} = \dfrac{1}{3}} \]