1. The problem is to calculate the combination $C(10,4)$, which represents the number of ways to choose 4 items from 10 without regard to order. 2. The formula for combinations is: $$C(n, r) = \frac{n!}{r!(n-r)!}$$ where $n!$ denotes the factorial of $n$, which is the product of all positive integers up to $n$. 3. Applying the formula to $C(10,4)$: $$C(10,4) = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!}$$ 4. Calculate the factorials or simplify by canceling common terms: $$10! = 10 \times 9 \times 8 \times 7 \times 6!$$ So, $$C(10,4) = \frac{10 \times 9 \times 8 \times 7 \times 6!}{4! \times 6!} = \frac{10 \times 9 \times 8 \times 7}{4!}$$ 5. Calculate $4!$: $$4! = 4 \times 3 \times 2 \times 1 = 24$$ 6. Substitute and compute: $$C(10,4) = \frac{10 \times 9 \times 8 \times 7}{24} = \frac{5040}{24} = 210$$ 7. Therefore, the number of ways to choose 4 items from 10 is $210$.