The set S = {1, 2, 3, ....... , 12} is to be partitioned into three sets A, B, C of equal size. Thus

Q: Wait, why do we not divide by $3!$ here? I thought when partitioning into groups of equal size, we always divide by the factorial of the number of groups.

That is a fantastic question! You are absolutely right that we divide by $3!$ when the groups are *unlabeled* or *indistinguishable*. However, in this problem, the sets are named $A, B,$ and $C$. This means that if we swap the contents of set $A$ and set $B$, we have created a different partition because the labels $A$ and $B$ are distinct. Therefore, the order of the groups matters, and we do not divide by $3!$.

Q: What if the sets were not of equal size? How would the formula change?

If the sets had different sizes, say $n_1, n_2,$ and $n_3$, the number of ways would simply be the product of the combinations: $\binom{n}{n_1} \times \binom{n-n_1}{n_2} \times \binom{n-n_1-n_2}{n_3}$. The division by $k!$ only comes into play when the groups are identical in size *and* indistinguishable in nature.

Q: Is there a faster way to think about this using the multinomial coefficient?

Yes! The multinomial coefficient $\frac{12!}{4!4!4!}$ directly counts the number of ways to arrange 12 items into three labeled boxes of size 4. It is mathematically equivalent to $\binom{12}{4} \times \binom{8}{4} \times \binom{4}{4}$, which we derived step-by-step. It is a powerful tool to keep in your arsenal!

Question

The set $S = \{1, 2, 3, \dots, 12\}$ is to be partitioned into three sets $A, B, C$ of equal size. Thus $A \cup B \cup C = S, A \cap B = B \cap C = A \cap C = \phi$. The number of ways to partition $S$ is

Options:

(A) $\frac{12!}{(4!)^3}$

(B) $\frac{12!}{(4!)^4}$

(C) $\frac{12!}{3!(4!)^3}$

(D) $\frac{12!}{3!(4!)^4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape! Today, we are embarking on a journey to solve a classic problem in combinatorics.

We are tasked with partitioning a set $S = \{1, 2, 3, \dots, 12\}$ into three distinct sets $A, B,$ and $C$, each containing exactly four elements. This is a gateway to understanding the profound difference between labeled and unlabeled groups, a concept that often trips up even the most seasoned JEE aspirants.

Imagine you are standing before three empty, distinct containers, each labeled with a unique name: $A, B,$ and $C$. You have twelve distinct items, and your goal is to distribute them such that each container holds exactly four items.

Because the containers are labeled, placing a specific set of four numbers into container $A$ is fundamentally different from placing that same set into container $B$. This distinction is the core of our problem.

### The Sequential Selection

Let us break this down into a sequence of deliberate choices. First, we approach container $A$. We have all twelve elements at our disposal and need to select four of them.

The number of ways to do this is given by the combination formula:

$$ \binom{12}{4} = \frac{12!}{4! \cdot 8!} $$

Now, with four elements safely tucked away in container $A$, we turn our attention to container $B$. We are left with $12 - 4 = 8$ elements. From these eight, we must choose four for container $B$:

$$ \binom{8}{4} = \frac{8!}{4! \cdot 4!} $$

Finally, we look at container $C$. We have only four elements remaining, and we must place all of them into this last container. The number of ways to do this is:

$$ \binom{4}{4} = 1 $$

### The Power of the Product Rule

Since these choices are sequential and independent, we apply the fundamental product rule of counting. The total number of ways to partition the set is the product of our individual choices:

$$ 	ext{Total Ways} = \binom{12}{4} 	imes \binom{8}{4} 	imes \binom{4}{4} $$

Substituting our factorial expressions, we get:

$$ 	ext{Total Ways} = \frac{12!}{4! \cdot 8!} 	imes \frac{8!}{4! \cdot 4!} 	imes \frac{4!}{4! \cdot 0!} $$

Observe the elegance of the cancellation! The $8!$ in the denominator of the first term cancels with the $8!$ in the numerator of the second. We are left with a beautifully simplified expression:

$$ 	ext{Total Ways} = \frac{12!}{4! \cdot 4! \cdot 4!} = \frac{12!}{(4!)^3} $$

### The Labeled vs. Unlabeled Trap

Here is the moment of truth. Many students, driven by habit, might be tempted to divide this result by $3!$.

That division is only necessary if the sets $A, B,$ and $C$ were indistinguishable—if they were just three identical, unlabeled piles. Because our sets are distinct, labeled entities, the order in which we fill them matters.

If we were to swap the contents of $A$ and $B$, we would have a different partition. Therefore, we do not divide by $3!$.

You have successfully navigated the trap! The final answer is:

$$ \mathbf{\frac{12!}{(4!)^3}} $$

Keep this logic close to your heart, and you will find that even the most complex combinatorial problems begin to yield to your understanding.

Animated Solution for Mathematics - Permutations and Combinations: The set $S = {1, 2, 3, \dots, 12}$ is to be partitioned into three sets $A, B, C$ of equal size. Thus $A \cup B \cup C = S, A \cap B = B \cap C = A \cap C = ϕ$ . The number of ways to partition $S$ is

Select Answer:

Visualized Solution

Visualizing the Set $S$

Understanding the Constraints

Filling the First Set $A$

Filling the Second Set $B$

Filling the Third Set $C$

Applying the Product Rule

Simplifying the Expression

Named vs. Unnamed Groups

The Sigma Insight: Formation of Groups

Analyzing the Setup

The Sequential Selection

The Power of the Product Rule

The Labeled vs

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