1. The problem is to calculate the combination $C(3,10)$, which represents the number of ways to choose 3 items from 10 without regard to order. 2. The formula for combinations is: $$C(n,r) = \frac{n!}{r!(n-r)!}$$ where $n!$ is the factorial of $n$, meaning the product of all positive integers up to $n$. 3. Applying the formula to $C(3,10)$: $$C(3,10) = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!}$$ 4. Calculate factorials: - $10! = 10 \times 9 \times 8 \times 7!$ - $3! = 3 \times 2 \times 1 = 6$ 5. Substitute and simplify: $$C(3,10) = \frac{10 \times 9 \times 8 \times 7!}{6 \times 7!} = \frac{10 \times 9 \times 8}{6}$$ 6. Simplify numerator and denominator: $$\frac{10 \times 9 \times 8}{6} = \frac{720}{6} = 120$$ 7. Therefore, the number of combinations is: $$C(3,10) = 120$$ This means there are 120 ways to choose 3 items from 10 without considering order.