Five balls of different colours are to be placed in there boxes of different size. Each box can hold

Q: Why do we divide by $2!$ when we have groups of size $2, 2, 1$? Isn't the selection unique?

Think of it this way: if you select balls $\{A, B\}$ for one group and $\{C, D\}$ for another, is that different from selecting $\{C, D\}$ first and $\{A, B\}$ second? No, the groups are just collections. Dividing by $2!$ removes the ordering of these identical-sized groups, ensuring we count the *partition* of the set, not the order in which we picked the groups.

Q: Can I use the Inclusion-Exclusion Principle instead of this case-by-case method?

Absolutely! It is a powerful tool. You would calculate $3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$. Both methods should yield the same result. The partitioning method is often more intuitive for visualizing the 'groups', while Inclusion-Exclusion is faster if you are comfortable with the formula.

Q: Why do we multiply by $3!$ at the end?

Because the boxes are distinct (different sizes). Once you have your three groups, you can permute them among the three boxes in $3!$ ways. If the boxes were identical, we wouldn't multiply by $3!$.

Question

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 ?

Sigma Percentile · Accepted Answer

### The Art of Distribution: Solving the Partition Puzzle

Welcome, future engineer! Today, we are diving into a classic problem that separates the casual student from the master of combinatorics. We have $5$ distinct balls and $3$ distinct boxes.

The constraint is simple yet powerful: no box can remain empty. This is not just a counting problem; it is a test of your ability to organize chaos into structure.

### Phase 1: The Philosophy of Partitioning

When you see a problem like this, do not rush to write down formulas. Instead, visualize the process. We have $5$ distinct balls to place into $3$ boxes.

If we just threw them in, we would have $3^5$ possibilities, but that includes scenarios where boxes are empty. To satisfy our constraint, we must first partition our $5$ balls into $3$ non-empty groups.

The only ways to write $5$ as a sum of $3$ positive integers are:

1. $3 + 1 + 1 = 5$
2. $2 + 2 + 1 = 5$

These are our two distinct scenarios. Let us tackle them one by one.

### Phase 2: Case 1 - The $(3, 1, 1)$ Split

Imagine we decide to put $3$ balls in one group and $1$ ball in each of the other two groups. We select $3$ balls from $5$ in $\binom{5}{3}$ ways, then $1$ from the remaining $2$ in $\binom{2}{1}$ ways, and the last ball in $\binom{1}{1}$ ways.

However, here lies the trap: the two groups of size $1$ are identical in size. If we simply multiply $\binom{5}{3} 	imes \binom{2}{1} 	imes \binom{1}{1}$, we overcount because the order of these two groups does not matter. We must divide by $2!$ to correct this.

The number of ways to form the groups is:

$$\frac{\binom{5}{3} 	imes \binom{2}{1} 	imes \binom{1}{1}}{2!} = \frac{10 	imes 2 	imes 1}{2} = 10$$

We have $10$ ways to form these groups. Since the boxes are distinct, we must multiply by $3!$ (which is $6$) to distribute these groups into the boxes. Thus, for Case 1, we have $10 	imes 6 = 60$ ways.

### Phase 3: Case 2 - The $(2, 2, 1)$ Split

Now, let us look at the second possibility: two groups of size $2$ and one group of size $1$. We select $2$ balls from $5$ in $\binom{5}{2}$ ways, then $2$ balls from the remaining $3$ in $\binom{3}{2}$ ways, and the last ball in $\binom{1}{1}$ ways.

Again, we see two groups of the same size ($2$). We must divide by $2!$ to avoid overcounting the identical groups.

$$\frac{\binom{5}{2} 	imes \binom{3}{2} 	imes \binom{1}{1}}{2!} = \frac{10 	imes 3 	imes 1}{2} = 15$$

We have $15$ ways to form these groups. Just like before, we distribute these $3$ groups into $3$ distinct boxes in $3!$ ways. So, for Case 2, we have $15 	imes 6 = 90$ ways.

### Phase 4: The Final Assembly

We have navigated the two possible worlds. Now, we simply bring them together. The total number of ways to distribute the balls such that no box is empty is the sum of our two cases:

$$	ext{Total Ways} = 60 + 90 = 150$$

There you have it! By breaking the problem into logical partitions and carefully handling the overcounting of identical group sizes, we have arrived at the solution.

Remember, combinatorics is not about memorizing formulas; it is about telling the story of how the objects are arranged. Keep practicing this level of logical rigor, and you will find that even the most complex JEE problems become a beautiful, solvable puzzle.

Animated Solution for Mathematics - Permutations and Combinations: 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 ?

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e P r o b l e m$

$I d e n t i f y in g t h e P oss ib l e G r o u p in g s$

$C a se 1 : F or min g G r o u p s o f 3, 1, 1$

$C a se 1 : C a l c u l a t in g G r o u p F or ma t i o n s$

$C a se 1 : D i s t r ib u t in g in t o B o x es$

$C a se 2 : F or min g G r o u p s o f 2, 2, 1$

$C a se 2 : C a l c u l a t in g G r o u p F or ma t i o n s$

$C a se 2 : D i s t r ib u t in g in t o B o x es$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Combinations and Selection

The Art of Distribution

Phase 1

Phase 2

Phase 3

Phase 4

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