1. **State the problem:** Calculate the combination with repetition for $n=28$ and $r=13$. 2. **Formula:** The formula for combinations with repetition is $$\binom{n+r-1}{r} = \frac{(n+r-1)!}{r! (n-1)!}$$ 3. **Substitute values:** Here, $n=28$ and $r=13$, so $$\binom{28+13-1}{13} = \binom{40}{13} = \frac{40!}{13! \times 27!}$$ 4. **Simplify:** This means we multiply the top 13 terms of 40! and divide by 13!: $$\frac{40 \times 39 \times 38 \times 37 \times 36 \times 35 \times 34 \times 33 \times 32 \times 31 \times 30 \times 29 \times 28}{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$$ 5. **Calculate:** Evaluating this expression gives the number of combinations with repetition. --- 1. **State the problem:** Calculate the combination with repetition for $n=15$ and $r=3$. 2. **Formula:** Using the same formula, $$\binom{15+3-1}{3} = \binom{17}{3} = \frac{17!}{3! \times 14!}$$ 3. **Simplify:** This is $$\frac{17 \times 16 \times 15}{3 \times 2 \times 1}$$ 4. **Calculate:** Simplify numerator and denominator: $$\frac{4080}{6} = 680$$ **Final answers:** - For $n=28$, $r=13$, the combination with repetition is $$\binom{40}{13} = \frac{40!}{13! \times 27!}$$ (exact value is large and typically computed with software). - For $n=15$, $r=3$, the combination with repetition is $$680$$.