Counting ways to split objects into three labeled groups

Expert intro: The multinomial coefficient counts labeled partitions of $n$ distinct objects into blocks of sizes $a$, $b$, and $c$. The product form of factorial ratios is the cleanest way to see why the final count is $\frac{n!}{a!\,b!\,c!}$ [1].

Detailed walkthrough

Core counting principle

When $n$ distinct objects are divided into three labeled groups of sizes $a$, $b$, and $c$ with $a+b+c=n$, the number of possible divisions is the multinomial coefficient $\frac{n!}{a!\,b!\,c!}.$ This counts how many ways we can choose which objects go into the first group, then the second, and finally the third.

A direct way to see it is to build the groups in sequence. First choose the $a$ objects for the first group: there are $\binom{n}{a}$ choices. From the remaining $n-a$ objects, choose $b$ for the second group: there are $\binom{n-a}{b}$ choices. The remaining $c$ objects must form the third group automatically. So the total number is $\binom{n}{a}\binom{n-a}{b} = \frac{n!}{a!(n-a)!}\cdot\frac{(n-a)!}{b!c!} = \frac{n!}{a!\,b!\,c!}.$

Why the factorial products match

The factorial formula in the prompt can be read as a chain of selections. The expression $\frac{n!}{(n-a)!}\cdot\frac{(n-a)!}{(n-a-b)!}\cdot\frac{(n-a-b)!}{(n-a-b-c)!}$ collapses by cancellation to $\frac{n!}{(n-a-b-c)!}.$ Because $a+b+c=n$, we have $n-a-b-c=0$, and since $0!=1$, this becomes $n!$ on top of the counting of ordered selections. To convert from ordered selections to unordered groups, we divide by $a!$, $b!$, and $c!$, because permutations inside each group do not create new groupings.

That is exactly why the final answer is $\frac{n!}{a!\,b!\,c!}$: the numerator counts all arrangements, while the denominator removes the internal reorderings that do not matter.

Common proof routes and hidden assumptions

A standard proof can be written in two equivalent ways. One method uses sequential choice and binomial coefficients, as above. Another uses permutation blocks: arrange all $n$ objects in a line, then cut the line into blocks of lengths $a$, $b$, and $c$. Since any order inside each block is irrelevant, divide by $a!b!c!$.

The key assumption is that the three groups are labeled or distinguished. If the groups were unlabeled, the count would be different when some of the group sizes are equal. Also, the objects must be distinct; otherwise the factorial counting would need adjustment for repeated items. For this question, distinct objects and fixed group sizes give the clean multinomial answer.

Common mistake to avoid

The biggest mistake is to treat the answer as merely $\binom{n}{a,b,c}$ without explaining where it comes from. In exam writing, you should show the chain of choices or the cancellation of factorials. Another common slip is forgetting that $a+b+c=n$; without that condition, the final cancellation does not reduce to a complete partition of all objects. The proof must explicitly use the fact that every object is assigned to exactly one of the three groups [2].

💡 Pitfall guide

The main place students lose marks in this multinomial proof is mixing up "choosing groups" with "ordering objects." A line such as $\frac{n!}{a!b!c!}$ may be correct, but it is not enough unless the reasoning shows why each denominator appears. The factor $a!$ removes permutations inside the first group, and similarly for $b!$ and $c!$. If you instead count by multiplying $n(n-1)\cdots$, you must be careful to stop at the correct stage and then divide out the internal rearrangements. Another frequent error is forgetting that the groups are labeled by size or position in the statement. If the problem says the objects are divided into groups of $a$, $b$, and $c$, then the groups are distinct in the counting process. If not, identical group sizes can lead to overcounting. Always check whether the statement wants an ordered distribution or just a partition.

🔄 Real-world variant

If the problem were changed to dividing $n$ distinct objects into four groups of sizes $a$, $b$, $c$, and $d$ with $a+b+c+d=n$, the same reasoning gives $\frac{n!}{a!\,b!\,c!\,d!}$. A proof would start by choosing the $a$ objects for the first group, then the $b$ for the second, then the $c$ for the third, leaving $d$ objects for the last group. The sequential choice formula becomes $\binom{n}{a}\binom{n-a}{b}\binom{n-a-b}{c}$, which simplifies to the four-part multinomial coefficient. This variant shows that the original argument is not special to three groups; it works for any fixed number of labeled groups as long as the sizes add up to $n$.

🔍 Related terms

multinomial coefficient, labeled partition count, sequential combinatorial choice