1. The problem is to evaluate the product of combinations: $\binom{6}{2} \times \binom{4}{2} \times \binom{2}{2}$. 2. Recall the formula for combinations: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ where $n!$ is the factorial of $n$. 3. Calculate each combination separately: - $\binom{6}{2} = \frac{6!}{2!4!} = \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} = \frac{30}{2} = 15$ - $\binom{4}{2} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} = \frac{12}{2} = 6$ - $\binom{2}{2} = \frac{2!}{2!0!} = \frac{2 \times 1}{2 \times 1 \times 1} = 1$ 4. Multiply the results: $$15 \times 6 \times 1 = 90$$ 5. Therefore, the value of $\binom{6}{2} \times \binom{4}{2} \times \binom{2}{2}$ is **90**.