1. The problem is to find the value of $C(8,5)$, which represents the number of combinations of 8 items taken 5 at a time. 2. The formula for combinations is: $$C(n,k) = \frac{n!}{k!(n-k)!}$$ where $n!$ is the factorial of $n$, and $k$ is the number of items chosen. 3. Substitute $n=8$ and $k=5$ into the formula: $$C(8,5) = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!}$$ 4. Calculate the factorials: $$8! = 8 \times 7 \times 6 \times 5!$$ So, $$C(8,5) = \frac{8 \times 7 \times 6 \times 5!}{5! \times 3!}$$ 5. Cancel the $5!$ terms: $$C(8,5) = \frac{8 \times 7 \times 6 \cancel{\times 5!}}{\cancel{5!} \times 3!}$$ 6. Calculate $3!$: $$3! = 3 \times 2 \times 1 = 6$$ 7. Substitute and simplify: $$C(8,5) = \frac{8 \times 7 \times 6}{6}$$ 8. Cancel the 6 in numerator and denominator: $$C(8,5) = \frac{8 \times 7 \times \cancel{6}}{\cancel{6}} = 8 \times 7 = 56$$ 9. Therefore, the number of combinations is: $$\boxed{56}$$