1. The problem asks to evaluate the combination $15C2$, which represents the number of ways to choose 2 items from 15 without regard to order. 2. The formula for combinations is: $$nCr = \frac{n!}{r!(n-r)!}$$ where $n$ is the total number of items, and $r$ is the number of items chosen. 3. Substitute $n=15$ and $r=2$: $$15C2 = \frac{15!}{2!(15-2)!} = \frac{15!}{2!13!}$$ 4. Simplify the factorial expression by canceling $13!$: $$15C2 = \frac{15 \times 14 \times \cancel{13!}}{2! \times \cancel{13!}} = \frac{15 \times 14}{2!}$$ 5. Calculate $2! = 2 \times 1 = 2$: $$15C2 = \frac{15 \times 14}{2}$$ 6. Perform the multiplication and division: $$15 \times 14 = 210$$ $$\frac{210}{2} = 105$$ 7. Since the result is an integer, the value of $15C2$ is 105. Final answer: $$15C2 = 105$$