1. **State the problem:** Calculate the combination $nCr$ for $n=5$ and $r=3$ using the formula for combinations. 2. **Formula:** The number of combinations of $n$ items taken $r$ at a time is given by: $$nCr = \frac{n!}{(n-r)!r!}$$ 3. **Apply the values:** Substitute $n=5$ and $r=3$: $$5C3 = \frac{5!}{(5-3)!3!} = \frac{5!}{2!3!}$$ 4. **Calculate factorials:** $$5! = 5 \times 4 \times 3 \times 2 \times 1$$ $$3! = 3 \times 2 \times 1$$ $$2! = 2 \times 1$$ 5. **Simplify numerator and denominator:** $$5C3 = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(3 \times 2 \times 1)}$$ 6. **Cancel common factors:** $$= \frac{5 \times 4 \times \cancel{3 \times 2 \times 1}}{(2 \times 1) \times \cancel{3 \times 2 \times 1}}$$ 7. **Simplify remaining terms:** $$= \frac{5 \times 4}{2 \times 1} = \frac{20}{2}$$ 8. **Final answer:** $$nCr = 10$$ This means there are 10 ways to choose 3 items from 5 without regard to order.