Distribution of 5 distinct balls into 3 distinct boxes — counting problems

Question

In how many ways can 5 distinct balls be distributed into 3 distinct boxes such that no box is empty?

(JEE Main 2023, similar pattern)

Solution — Step by Step

Each of the 5 balls can go into any of the 3 boxes independently. So the total number of ways (with no restriction):

\text{Total} = 3^5 = 243

Let $A_i$ = set of distributions where box $i$ is empty.

We want: Total $- |A_1 \cup A_2 \cup A_3|$

By inclusion-exclusion:

|A_1 \cup A_2 \cup A_3| = \sum|A_i| - \sum|A_i \cap A_j| + |A_1 \cap A_2 \cap A_3|

$|A_i|$ = distributions with one specific box empty = $2^5 = 32$ (5 balls into remaining 2 boxes)

There are $\binom{3}{1} = 3$ such terms: $3 \times 32 = 96$

$|A_i \cap A_j|$ = distributions with two specific boxes empty = $1^5 = 1$ (all balls in the remaining 1 box)

There are $\binom{3}{2} = 3$ such terms: $3 \times 1 = 3$

$|A_1 \cap A_2 \cap A_3|$ = all three boxes empty = $0$ (impossible, balls must go somewhere)

|A_1 \cup A_2 \cup A_3| = 96 - 3 + 0 = 93

Number of distributions with no box empty:

243 - 93 = \boxed{150}

Why This Works

The inclusion-exclusion principle systematically removes overcounted cases. First, we subtract all distributions where at least one box is empty. But this double-subtracts distributions where two boxes are empty, so we add those back. The formula ensures exact counting.

This is actually a specific case of the Surjective function count: the number of onto functions from a set of 5 elements to a set of 3 elements. The general formula for surjections from $n$ elements to $r$ elements is:

S(n, r) = \sum_{k=0}^{r} (-1)^k \binom{r}{k}(r-k)^n

For $n = 5, r = 3$ : $\binom{3}{0}(3)^5 - \binom{3}{1}(2)^5 + \binom{3}{2}(1)^5 = 243 - 96 + 3 = 150$ .

Alternative Method — Using Stirling numbers

The number of distributions = $r! \times S(n, r)$ where $S(n, r)$ is the Stirling number of the second kind.

$S(5, 3) = 25$ (number of ways to partition 5 objects into 3 non-empty groups).

Since the boxes are distinct: $3! \times 25 = 6 \times 25 = 150$ .

For JEE, the distribution formula with inclusion-exclusion is the most reliable method. For small numbers, you can also enumerate: the possible distributions of 5 balls into 3 non-empty boxes are (3,1,1), (2,2,1) types. Count each separately. But for larger numbers, inclusion-exclusion scales better.

Common Mistake

Students sometimes try to split the problem as “choose 3 balls for 3 boxes, then distribute the remaining 2.” This leads to complicated casework and double-counting. The inclusion-exclusion approach is cleaner: start with all $3^5$ distributions and subtract the ones with empty boxes. Don’t overcomplicate it with manual casework.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Stirling numbers

Common Mistake

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