1. The problem asks to find the value of $\frac{\binom{9}{6}}{\binom{6}{2}}$. 2. Recall the combination formula: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$, where $n!$ is the factorial of $n$. 3. Calculate $\binom{9}{6}$: $$\binom{9}{6} = \frac{9!}{6!3!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$ 4. Calculate $\binom{6}{2}$: $$\binom{6}{2} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$$ 5. Now divide the two results: $$\frac{\binom{9}{6}}{\binom{6}{2}} = \frac{84}{15} = 5.6$$ 6. Therefore, the answer is 5.6.