The set S : {1, 2, 3, ...., 12} is to be partitioned into three sets A, B, C of equal size. Thus A B C = S, A B = B C = A C = . The number of ways to partition S is-

1. Core Concept – Multinomial Coefficient

When we split $n$ distinct objects into groups of fixed sizes $n_1, n_2, …, n_k$ and we care about the labels of the groups, the count is the multinomial coefficient

$\frac{n!}{n_1!\,n_2!\,\dots\,n_k!}$

Here, $n = 12$ and each $n_i = 4$. So, if the groups A, B, C were labelled (order matters), we would have

$\frac{12!}{4!\,4!\,4!}$

2. Removing Over-counting – Groups Are Unlabelled

In the problem, the sets A, B, C are just three piles—interchanging their names makes no new partition. There are $3!$ possible ways to permute (rename) the three groups, so we must divide by $3!$ to avoid counting the same split multiple times.

3. Logical Steps a Student Would Take

1. Select A: Choose any 4 numbers from the 12 ⇒ ${12\choose4}$.
2. Select B: Choose 4 of the remaining 8 ⇒ ${8\choose4}$.
3. Select C: The last 4 are forced ⇒ ${4\choose4}=1$.
4. Unlabelled adjustment: Divide by $3!$ because (A,B,C) is the same as (B,C,A), etc.

Putting it together:

$\text{Total ways} = \frac{1}{3!}\,{12\choose4}{8\choose4}{4\choose4}$

Imagine Sharing 12 Candies with 3 Friends

You have 12 different candies, and you want to give exactly 4 candies to each of your 3 friends (let’s call them Friend-1, Friend-2, Friend-3).
But in the end you do not care which friend gets which 4 candies – you only care that the candies are split into three piles of 4.
The question is: In how many different ways can you make those three piles?

So it’s just like cutting the big set {1, 2, …, 12} into three mini-sets of size 4 each. You’re counting all the possible fair splits!

Step-by-Step Solution

1. Labelled split
   $\frac{12!}{4!\,4!\,4!}$
2. Unlabelled correction
   Divide by $3!$.

Hence