The total number of ways in which 5 balls of different colours can be distributed among 3 persons so

Q: Wait, why can't we just give one ball to each person first and then distribute the remaining two?

That is a common trap! If you give one ball to each person first, you are assuming the balls are identical or that the order of distribution doesn't matter in a specific way. Because the balls are distinct, the number of ways to choose which ball goes to which person matters. This method leads to overcounting or undercounting because it doesn't account for all permutations of the balls.

Q: Why do we add back the cases where two people are empty? Shouldn't we just subtract the empty cases?

Think of it like a Venn diagram. When you subtract the cases where Person 1 is empty, you also subtract cases where Person 1 AND Person 2 are empty. When you subtract the cases where Person 2 is empty, you subtract the cases where Person 1 AND Person 2 are empty again. You've subtracted those 'double-empty' cases twice! Adding them back once balances the equation perfectly.

Q: Is there a faster way to do this without PIE?

Yes! You can use Stirling Numbers of the Second Kind, denoted as $S(n, k)$, which counts the ways to partition $n$ distinct objects into $k$ non-empty identical groups. Since our people are distinct, we multiply by $k!$. So, the answer is $3! \times S(5, 3)$. It is a very elegant shortcut once you master the concept!

Question

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

Options:

(A) 75

(B) 150

(C) 210

(D) 243

Sigma Percentile · Accepted Answer

### The Art of Distribution: A Combinatorial Journey

Welcome, fellow explorer of the mathematical universe! Today, we are tackling a classic problem that sits at the heart of combinatorics: distributing distinct items into distinct containers with a strict constraint.

Imagine you have $5$ vibrant, distinct balls—perhaps red, blue, green, yellow, and purple—and you want to share them among $3$ friends. The catch? Every single friend must receive at least one ball. No one can be left empty-handed.

### Phase 1: The Unrestricted Universe

Before we worry about the constraint, let's imagine a world where anything goes. If we ignore the 'at least one' rule, how many ways can we distribute these $5$ balls?

Each ball is an independent agent. The red ball has $3$ choices (Friend 1, 2, or 3). The blue ball also has $3$ choices, and so on for all $5$ balls.

By the fundamental multiplication principle, the total number of ways is:

$$3 	imes 3 	imes 3 	imes 3 	imes 3 = 3^5 = 243$$

This is our 'unrestricted universe.' It contains every possible outcome, including the 'bad' ones where someone gets nothing.

### Phase 2: The Inclusion-Exclusion Strategy

Now, we must prune our universe. We need to remove the scenarios where at least one person is empty. This is where the **Principle of Inclusion-Exclusion (PIE)** becomes our best friend.

We start with our total, $N = 243$. We want to subtract the cases where at least one person is empty. Let $S_1$ be the number of ways where at least one person is empty.

To find $S_1$, we choose one person to be empty in $\binom{3}{1}$ ways. The remaining $5$ balls must then be distributed among the remaining $2$ people. Each ball now has only $2$ choices, so there are $2^5$ ways.

$$S_1 = \binom{3}{1} 	imes 2^5 = 3 	imes 32 = 96$$

But wait! By subtracting $S_1$, we have over-subtracted. Specifically, the cases where two people are empty were subtracted twice (once for each empty person). We must add these back.

Let $S_2$ be the number of ways where at least two people are empty. We choose $2$ people to be empty in $\binom{3}{2}$ ways. The remaining $5$ balls must all go to the one remaining person, which is $1^5$ way.

$$S_2 = \binom{3}{2} 	imes 1^5 = 3 	imes 1 = 3$$

Our final formula is:

$$	ext{Valid Ways} = N - S_1 + S_2$$

### Phase 3: The Elegant Conclusion

Putting it all together: $243 - 96 + 3$. First, $243 - 96 = 147$. Then, adding back the $3$ gives us **$150$**.

It is a beautiful, clean result. We have successfully navigated the constraints and found the exact number of valid distributions.

Remember, whenever you face a problem with 'at least one' constraints, let PIE be your guiding light. And for those who love patterns, remember that this is also $3! 	imes S(5, 3)$, where $S(5, 3)$ is the Stirling number of the second kind. Keep practicing, keep visualizing, and most importantly, keep falling in love with the logic behind the numbers!

Animated Solution for Mathematics - Permutations and Combinations: The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

Select Answer:

Visualized Solution

Visualizing the Distribution

Total Unrestricted Ways

Calculating $3^{5}$

The Inclusion-Exclusion Strategy

Case: At least one person empty ( $S_{1}$ )

Calculating $S_{1}$

Case: At least two persons empty ( $S_{2}$ )

Calculating $S_{2}$

Final Computation - Part 1

Final Computation - Part 2

Key Takeaway & Summary

The Sigma Insight: Combinations and Selection

The Art of Distribution

Phase 1

Phase 2

Phase 3

Similar Questions

Five balls of different colours are to be placed in there boxes of different size. Each box can hold all five. In how many different ways can we place the balls so that no box remains empty ?

The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is _____

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is :

The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is

The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is ....

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at most three of them are red is __________.