To solve this problem, we need to divide 12 presents into 4 piles, with each pile containing 3 presents. The key point here is that the order of the piles does not matter (i.e., the piles are indistinguishable). So, we are looking for the number of distinct ways to partition the 12 presents into 4 groups of 3.

Steps to solve:

1. Total number of ways to divide without considering identical piles:

   First, we can think of this as arranging the 12 presents in a sequence and then dividing them into groups. The total number of ways to choose 3 presents for the first pile, 3 for the second, and so on is: $\frac{12!}{(3!)^4}$ Here:

   • $12!$ is the total number of ways to arrange 12 presents.
   • $(3!)^4$ accounts for the fact that within each pile, the order of presents doesn't matter (because the piles are indistinguishable).

   Calculating this step-by-step: $12! = 479,001,600$ $(3!)^4 = 6^4 = 1,296$

   So, the total number of ways: $\frac{12!}{(3!)^4} = \frac{479,001,600}{1,296} = 369,600$

2. Divide by the number of ways to rearrange the 4 piles (since the piles are indistinguishable):

   Since the piles are indistinguishable, we also need to divide by the number of ways to arrange the 4 piles, which is $4!$: $4! = 24$

   Thus, the total number of distinct ways to divide the presents is: $\frac{369,600}{24} = 15,400$

Conclusion:

There are 15,400 distinct ways to divide 12 presents into 4 piles of 3 presents.

Would you like further explanation, or do you have any questions?

Here are 5 related questions:

1. How many ways can 15 presents be divided into 5 piles of 3?
2. How does this approach change if the piles are distinguishable?
3. What if we had 12 presents and needed 3 piles of 4 presents?
4. How does this problem relate to combinations and partitions?
5. How would this change if there were constraints on which presents can go into certain piles?

Tip: When dividing objects into groups, always check if the groups are distinguishable or not. This significantly affects the formula.