How many ways to put 10 IDENTICAL balls into 4 DISTINCT boxes (each box can hold 0 or more)?

1. STARS AND BARS: count solutions to $x_1 + x_2 + x_3 + x_4 = 10$ where $x_i \geq 0$. 2. Formula: number of non-negative integer solutions to $\sum x_i = n$ with $k$ variables is $\binom{n+k-1}{k-1}$. 3. Here $n = 10$, $k = 4$: $\binom{10+4-1}{4-1} = \binom{13}{3} = \dfrac{13!}{3!\,10!} = \dfrac{13 \cdot 12 \cdot 11}{6} = 286$. 4. Intuition: 10 stars + 3 bars to delimit 4 groups; arrange them in $\binom{13}{3}$ ways. 5. Option A is the wrong formula. Option B treats balls as DISTINCT (wrong). Option D is for ordered DISTINCT items. _Source: Oscar Levin, "Discrete Mathematics: An Open Introduction", §1.5 (Stars and Bars)._