1. The problem is to find the value of the binomial coefficient $\binom{9}{8}$. 2. The formula for a binomial coefficient is: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ where $n!$ means factorial of $n$, which is the product of all positive integers up to $n$. 3. Substitute $n=9$ and $k=8$ into the formula: $$\binom{9}{8} = \frac{9!}{8!(9-8)!} = \frac{9!}{8!1!}$$ 4. Simplify the factorial expressions: $$9! = 9 \times 8!$$ So, $$\binom{9}{8} = \frac{9 \times 8!}{8! \times 1}$$ 5. Cancel the common factor $8!$: $$\binom{9}{8} = \frac{9 \times \cancel{8!}}{\cancel{8!} \times 1} = 9$$ 6. Therefore, the value of $\binom{9}{8}$ is 9. This means there are 9 ways to choose 8 items from 9 items.