1. The problem asks to evaluate the combination $14C5$, which represents the number of ways to choose 5 items from 14 without regard to order. 2. The formula for combinations is: $$nCr = \frac{n!}{r!(n-r)!}$$ where $n = 14$ and $r = 5$. 3. Substitute the values: $$14C5 = \frac{14!}{5!(14-5)!} = \frac{14!}{5!9!}$$ 4. Expand factorials to simplify: $$14! = 14 \times 13 \times 12 \times 11 \times 10 \times 9!$$ So, $$14C5 = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9!}{5! \times 9!}$$ 5. Cancel $9!$ from numerator and denominator: $$14C5 = \frac{14 \times 13 \times 12 \times 11 \times 10}{5!}$$ 6. Calculate $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$. 7. Calculate numerator: $$14 \times 13 = 182$$ $$182 \times 12 = 2184$$ $$2184 \times 11 = 24024$$ $$24024 \times 10 = 240240$$ 8. Divide numerator by denominator: $$14C5 = \frac{240240}{120}$$ 9. Simplify the fraction by canceling common factors: $$\frac{\cancel{240240}}{\cancel{120}} = 2002$$ 10. The value is an integer. Final answer: $$14C5 = 2002$$