Dividing Students into Labeled Groups
A class of 12 students must split into 4 groups of 3, where the groups are labeled 1 through 4 (i.e., the groups are distinguishable). How many distinct ways can the class form these groups?
Hints
- Since the groups are labeled (numbered 1-4), assigning different students to different group numbers counts as a distinct arrangement.
- Think of each assignment as an anagram of a 12-character string with 3 copies each of the digits 1, 2, 3, 4.
- The count is the multinomial coefficient $\binom{12}{3,3,3,3} = \frac{12!}{(3!)^4}$.
Worked Solution
How to Think About It: The key question in any "partition into groups" problem is: are the groups distinguishable or not? Here, the groups are numbered 1-4, so they are distinguishable. That makes things simpler -- assigning student A to Group 1 and student B to Group 2 is different from assigning A to Group 2 and B to Group 1. If the groups were unlabeled, you would need to divide by the number of ways to permute the groups, but that is not the case here.
Quick Estimate: We are distributing 12 people into 4 labeled bins of size 3. A rough sanity check: there are $12!$ ways to line up all 12 students, and we need to divide by $3!$ for the ordering within each group (since groups are sets, not sequences). That gives $12! / (3!)^4$. With $12! = 479{,}001{,}600$ and $(3!)^4 = 1296$, we get about 370,000. Let's verify.
Approach: This is a multinomial coefficient problem.
Formal Solution:
Assigning 12 distinct students to 4 labeled groups of size 3 is equivalent to counting anagrams of a string with 3 copies each of 4 distinct symbols (say, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4). Each arrangement corresponds to a group assignment: if the $i$-th character is $j$, student $i$ goes to group $j$.
The number of such anagrams is the multinomial coefficient:
$$\binom{12}{3, 3, 3, 3} = \frac{12!}{3! \cdot 3! \cdot 3! \cdot 3!} = \frac{12!}{(3!)^4}$$
Computing: $$\frac{479{,}001{,}600}{6^4} = \frac{479{,}001{,}600}{1{,}296} = 369{,}600$$
Alternatively, think of it sequentially: choose 3 of 12 for group 1 ($\binom{12}{3}$), then 3 of the remaining 9 for group 2 ($\binom{9}{3}$), then 3 of 6 for group 3 ($\binom{6}{3}$), and the last 3 go to group 4 ($\binom{3}{3}$):
$$\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3} = 220 \times 84 \times 20 \times 1 = 369{,}600$$
Answer: There are $\dfrac{12!}{(3!)^4} = 369{,}600$ distinct ways.
Intuition
Partition-into-groups problems come in two flavors: labeled groups and unlabeled groups. The labeled version (this problem) is simpler because every assignment of people to group numbers is distinct. The unlabeled version requires dividing by $k!$ to account for the fact that swapping entire groups does not create a new partition. Always ask yourself: does it matter which group is which? If the groups have names, roles, or numbers, they are labeled. If they are interchangeable ("just divide into teams of 3"), they are unlabeled.
This distinction appears in quantitative problems more often than you might expect. In combinatorics interview questions, the most common mistake is forgetting to divide by $k!$ for unlabeled groups (or dividing when you should not). The multinomial coefficient $\binom{n}{n_1, n_2, \ldots, n_k}$ is the workhorse formula for distributing distinct objects into labeled bins of specified sizes.