Find number of ways to distribute 10 identical balls in 3 distinct boxes

Question

Find the number of ways to distribute 10 identical balls into 3 distinct boxes (where boxes may be empty).

Solution — Step by Step

This is a stars and bars (combinations with repetition) problem. We’re distributing identical objects into distinct containers — the key word “identical” means we can’t distinguish between balls, only between how many go into each box.

If the balls were distinct, the answer would be $3^{10}$ (each ball independently chooses one of 3 boxes). But since balls are identical, we only care about how many go into each box, not which specific ball.

Let $x_1$ , $x_2$ , $x_3$ be the number of balls in Box 1, Box 2, Box 3 respectively. We need:

x_1 + x_2 + x_3 = 10, \quad x_i \geq 0

The number of non-negative integer solutions to this equation counts exactly the number of ways to distribute the balls.

The number of non-negative integer solutions to $x_1 + x_2 + ... + x_r = n$ is:

\binom{n + r - 1}{r - 1}

Here $n = 10$ (balls) and $r = 3$ (boxes):

\binom{10 + 3 - 1}{3 - 1} = \binom{12}{2}

\binom{12}{2} = \frac{12 \times 11}{2} = \frac{132}{2} = \mathbf{66}

There are 66 ways to distribute 10 identical balls into 3 distinct boxes.

Why This Works

The “stars and bars” model: imagine 10 stars (representing balls) and 2 bars (representing dividers between 3 boxes). Any arrangement of these 12 symbols (10 stars + 2 bars) corresponds to a valid distribution. For example, ★★★|★★★★★|★★ means 3 balls in Box 1, 5 in Box 2, 2 in Box 3.

We need to choose 2 positions (out of 12) for the bars, or equivalently, choose 2 positions for the dividers among the 12 total positions. That’s $\binom{12}{2} = 66$ .

Alternative Method

Systematic listing (works for small cases, not practical for 10 balls):

For 3 balls and 3 boxes: solutions to $x_1 + x_2 + x_3 = 3$ include (3,0,0), (0,3,0), (0,0,3), (2,1,0), (2,0,1), (1,2,0), (0,2,1), (1,0,2), (0,1,2), (1,1,1) = 10 solutions.

Check formula: $\binom{3+3-1}{3-1} = \binom{5}{2} = 10$ ✓

For JEE and CBSE Class 11, this formula is tested both directly and as a component of larger problems. The condition matters: if boxes cannot be empty, use $\binom{n-1}{r-1}$ (subtract 1 ball per box first, then distribute the rest). For the “at least 1 ball in each box” version here: distribute 7 balls freely = $\binom{7+2}{2} = \binom{9}{2} = 36$ .

Common Mistake

The most common confusion is between identical and distinct objects. If the 10 balls were distinct (like coloured balls), the answer would be $3^{10} = 59049$ — each ball independently chooses a box. The Stars and Bars approach applies ONLY when objects are identical (indistinguishable). If you see “identical balls/coins/objects,” use $\binom{n+r-1}{r-1}$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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