Question

If \( ^{2n}C_3 : ^nC_3 = 12 : 1 \), then \( n = \) ?

• A. 5
• B. 8
• C. 10
• D. 3

Hint:

Use combination formulas \( ^nC_r = \frac{n(n-1)...(n-r+1)}{r!} \) for simplifying ratios. Cancel factorials when possible.

Answer 1

The Correct Option is A

We are given: \[ \frac{{^{2n}C_3}}{{^nC_3}} = 12 \] Use the combination formula: \[ ^rC_3 = \frac{r(r - 1)(r - 2)}{6} \] So, \[ \frac{\frac{2n(2n - 1)(2n - 2)}{6}}{\frac{n(n - 1)(n - 2)}{6}} = 12 \Rightarrow \frac{2n(2n - 1)(2n - 2)}{n(n - 1)(n - 2)} = 12 \] Simplify numerator: \[ 2n(2n - 1)(2n - 2) = 2n(2n - 1)(2(n - 1)) = 4n(2n - 1)(n - 1) \] So: \[ \begin{align} \frac{4n(2n - 1)(n - 1)}{n(n - 1)(n - 2)} = 12 \Rightarrow \frac{4(2n - 1)}{(n - 2)} = 12 \Rightarrow 4(2n - 1) = 12(n - 2) \Rightarrow 8n - 4 = 12n - 24 \Rightarrow 4n = 20 \Rightarrow n = 5 \]