1. **State the problem:** Calculate the combination formula given by $$\frac{(n+r-1)!}{r!(n-1)!}$$ for the values $n=35$ and $r=10$. 2. **Formula used:** The formula for combinations with repetition is: $$C(n+r-1, r) = \frac{(n+r-1)!}{r!(n-1)!}$$ This formula counts the number of ways to choose $r$ elements from $n$ types with repetition allowed. 3. **Substitute the values:** $$\frac{(35+10-1)!}{10!(35-1)!} = \frac{44!}{10! \times 34!}$$ 4. **Simplify the factorial expression:** Expand the numerator from $44!$ down to $35!$ to cancel with the denominator $34!$: $$\frac{44 \times 43 \times 42 \times 41 \times 40 \times 39 \times 38 \times 37 \times 36 \times 35 \times 34!}{10! \times 34!} = \frac{44 \times 43 \times 42 \times 41 \times 40 \times 39 \times 38 \times 37 \times 36 \times 35}{10!}$$ 5. **Calculate the denominator:** $$10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3628800$$ 6. **Calculate the numerator:** Multiply the numbers: $$44 \times 43 \times 42 \times 41 \times 40 \times 39 \times 38 \times 37 \times 36 \times 35 = 472972032000$$ 7. **Divide numerator by denominator:** $$\frac{472972032000}{3628800} = 13032096$$ **Final answer:** $$\boxed{13032096}$$