Distribution of Balls in Boxes – 4 Standard Cases (with Worked Examples)

We study 4 classic cases:

1. Different balls, different boxes (DB–DB)
2. Identical balls, different boxes (IB–DB)
3. Different balls, identical boxes (DB–IB)
4. Identical balls, identical boxes (IB–IB)

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1. Different Balls & Different Boxes (DB–DB)

Example 1

Problem: 3 distinct balls $A,B,C$ into 2 distinct boxes $X,Y$, no restriction. How many ways?

Solution: Each ball has 2 choices (X or Y). So total ways $= 2^3 = 8$.

Example 2

Problem: 3 distinct balls $A,B,C$ into 3 distinct boxes $X,Y,Z$, no box empty.

Solution: Total functions from 3 balls to 3 boxes $= 3^3 = 27$. Subtract distributions where some box is empty (onto counting):

• Exactly 1 box empty: choose the empty box in $^3C_1 = 3$ ways, remaining 2 boxes get all 3 balls: $2^3 = 8$ ways. Total $= 3 \times 8 = 24$.

Distributions with at least one empty box $= 24$. So onto (no box empty) $= 27 - 24 = 3$ ways.

Example 3

Problem: 4 distinct balls into 2 distinct boxes, each box must get at least 1 ball.

Solution: Total ways (no restriction) $= 2^4 = 16$. Cases with some box empty: if box X empty, all 4 in Y (1 way); if box Y empty, all 4 in X (1 way). So invalid $= 2$ ways.

Valid ways $= 16 - 2 = 14$.

Example 4

Problem: 4 distinct balls and 4 distinct boxes, each box gets exactly one ball.

Solution: This is just permutations of 4 balls into 4 boxes: $4! = 24$.

Example 5

Problem: 5 distinct balls, 3 distinct boxes, each box can hold any number of balls. How many distributions?

Solution: Each ball has 3 choices (which box). Total ways $= 3^5 = 243$.

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2. Identical Balls & Different Boxes (IB–DB)

Use stars and bars for equations like $x_1 + x_2 + \dots + x_k = N$.

Example 1

Problem: 4 identical balls into 3 distinct boxes, no restriction.

Solution: Let $x_1,x_2,x_3$ be number of balls in each box.

$x_1 + x_2 + x_3 = 4$, $x_i \ge 0$. Number of non-negative solutions $= {}^{4+3-1}C_{3-1} = {}^6C_2 = 15$.

Example 2

Problem: 5 identical balls into 3 distinct boxes, each box must get at least 1 ball.

Solution: Let $x_i \ge 1$ be balls per box, $x_1 + x_2 + x_3 = 5$.

Convert: $x_i = y_i + 1$, $y_i \ge 0$.

$y_1 + y_2 + y_3 = 5 - 3 = 2$. Number of solutions $= {}^{2+3-1}C_{3-1} = {}^4C_2 = 6$.

Example 3

Problem: 6 identical balls into 4 distinct boxes, first box must contain at least 2 balls.

Solution: Let box1 contain $x_1 \ge 2$, others $x_2,x_3,x_4 \ge 0$.

$x_1 + x_2 + x_3 + x_4 = 6$.

Put $x_1 = y_1 + 2$, $y_1 \ge 0$:

$y_1 + x_2 + x_3 + x_4 = 4$ with all $\ge 0$. Solutions $= {}^{4+4-1}C_{4-1} = {}^7C_3 = 35$.

Example 4

Problem: 7 identical balls into 3 distinct boxes, each box contains at most 4 balls.

Solution: Let $x_1+x_2+x_3=7$, $0 \le x_i \le 4$.

Total non-negative solutions (no upper bound) $= {}^{7+3-1}C_{3-1} = {}^9C_2 = 36$.

Subtract cases where some $x_i \ge 5$. By symmetry, count for $x_1 \ge 5$ and multiply by 3.

Let $x_1 = 5 + y_1$, $y_1 \ge 0$:

$y_1 + x_2 + x_3 = 2$ → ${}^{2+3-1}C_{3-1} = {}^4C_2 = 6$.

For $x_1 \ge 5$ we get 6 solutions; same for $x_2$ and $x_3$ → $3 \times 6 = 18$.

No overlap possible because sum is only 7 (can’t have two variables ≥5). So valid solutions $= 36 - 18 = 18$.

Example 5

Problem: 8 identical balls into 4 distinct boxes, at least one box empty.

Solution: Total non-negative solutions of $x_1+x_2+x_3+x_4=8$:

${}^{8+4-1}C_{4-1} = {}^{11}C_3 = 165$.

Solutions with no box empty (all $x_i \ge 1$): Put $x_i = y_i + 1$ → $y_1 + y_2 + y_3 + y_4 = 4$:

${}^{4+4-1}C_{4-1} = {}^7C_3 = 35$.

At least one box empty $= 165 - 35 = 130$.

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3. Different Balls & Identical Boxes (DB–IB)

Boxes are not labeled. Only the grouping (which balls together) matters. We count partitions of the set of balls, taking care of equal-size groups.

Example 1

Problem: 3 distinct balls $\{A,B,C\}$ into 2 identical boxes, boxes can be empty.

Solution: “Two identical boxes” but allowing empties is same as deciding which balls go together, because empty box doesn’t add a new pattern. So we are just partitioning the set into at most 2 groups.

• All 3 together in one box: $\{A,B,C\}$.
• Split as 2+1: $\{A,B\}\{C\}$, $\{A,C\}\{B\}$, $\{B,C\}\{A\}$.

Total ways $= 4$.

Example 2

Problem: 3 distinct balls into 3 identical boxes, no box empty.

Solution: If all 3 boxes are non-empty with identical boxes, each box must get exactly 1 ball.

But boxes are identical, so any “who goes to which box” looks the same: each ball in its own box. So only 1 way.

Example 3

Problem: 4 distinct balls $\{A,B,C,D\}$ into 2 identical boxes, no box empty.

Solution: We split 4 balls into 2 unlabeled groups, both non-empty.

Possible size patterns: 1+3 or 2+2.

• 1+3 split: choose the 1-ball group in $^4C_1 = 4$ ways. The other 3 go together. Since boxes are identical, 1+3 and 3+1 are same, so these 4 are all distinct partitions.
• 2+2 split: choose 2 balls for first group: $^4C_2 = 6$, but each partition is counted twice (group1/group2 swapped), so distinct partitions $= 6/2 = 3$.

Total ways $= 4 + 3 = 7$.

Example 4

Problem: 4 distinct balls into 3 identical boxes, any box may be empty.

Solution: We only care about how balls are grouped.

Possible group size patterns (allow ≤3 groups):

• 4 (all in one box): 1 way: $\{A,B,C,D\}$.
• 3+1: choose 1 alone: $^4C_1 = 4$ partitions.
• 2+2: as in Example 3 → $3$ partitions.
• 2+1+1: choose pair first: $^4C_2 = 6$, the remaining two singles are identical boxes, but singles are just individual balls, so no division by 2. So $6$ partitions.
• 1+1+1+1 would need 4 boxes, but we only have 3 boxes, so not allowed.

Total ways $= 1 + 4 + 3 + 6 = 14$.

Example 5

Problem: 5 distinct balls into 2 identical boxes, empty box allowed.

Solution: We partition 5 balls into ≤2 unlabeled groups.

Patterns: all 5 together (5), or 4+1, or 3+2.

• 5: $\{A,B,C,D,E\}$ → 1 way.
• 4+1: choose the single ball: $^5C_1 = 5$ ways.
• 3+2: choose the 3-ball group: $^5C_3 = 10$, but 3+2 and 2+3 are same (boxes identical), so distinct partitions $= 10/2 = 5$.

Total ways $= 1 + 5 + 5 = 11$.

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4. Identical Balls & Identical Boxes (IB–IB)

Now only the pattern “how many balls in each box” matters, boxes are not labeled, balls are not labeled. This is essentially integer partition with at most $k$ parts.

Example 1

Problem: 4 identical balls into 2 identical boxes, boxes may be empty.

Solution: We need unordered pairs $(a,b)$ with $a+b=4$, $a,b \ge 0$, and $(a,b)$ ≡ $(b,a)$.

Possible $(a,b)$: (4,0), (3,1), (2,2). So there are 3 distinct ways.

Example 2

Problem: 5 identical balls into at most 3 identical boxes, boxes may be empty.

Solution: We want partitions of 5 into ≤3 parts:

• 5
• 4 + 1
• 3 + 2
• 3 + 1 + 1
• 2 + 2 + 1

That is 5 ways.

Example 3

Problem: 6 identical balls into 3 identical boxes, no box empty.

Solution: Partitions of 6 into exactly 3 positive parts (order not important):

• 4 + 1 + 1
• 3 + 2 + 1
• 2 + 2 + 2

So there are 3 ways.

Example 4

Problem: 7 identical balls into at most 3 identical boxes, boxes may be empty.

Solution: Partitions of 7 into ≤3 parts:

• 7
• 6 + 1
• 5 + 2
• 5 + 1 + 1
• 4 + 3
• 4 + 2 + 1
• 3 + 3 + 1
• 3 + 2 + 2

Total ways $= 8$.

Example 5

Problem: 8 identical balls into exactly 3 identical boxes, boxes non-empty.

Solution: Partitions of 8 into exactly 3 positive parts:

• 6 + 1 + 1
• 5 + 2 + 1
• 4 + 3 + 1
• 4 + 2 + 2
• 3 + 3 + 2

So there are 5 ways.

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These 20 examples cover all four standard models of balls-and-boxes questions and are ideal for building concept + question bank for NIMCET / CUET / SSC etc.