Multinomial Coefficients

Count the ways to partition a set into multiple labeled groups of specified sizes.

1. The idea

The multinomial coefficient generalizes combinations to more than two groups: the number of ways to split $n$ distinct items into labeled groups of sizes $n_{1}, n_{2}, \dots, n_{k}$ (with $n_{1} + \dots + n_{k} = n$ ) is $\frac{n !}{n _{1} ! n _{2} ! \dots n _{k} !}$ .

You can derive this by sequential combinations: choose $n_{1}$ items for group 1 from $n$ , then $n_{2}$ from the remaining $n - n_{1}$ for group 2, and so on. The product of binomial coefficients telescopes into the single multinomial formula.

Multinomial coefficients also count arrangements of a multiset — for example, the number of distinct orderings of the letters in a word with repeated letters, where each repeated letter's factorial is divided out.

2. Why it matters

Splitting people or items into multiple distinguishable groups is a very common interview framing (teams, shifts, categories), and the multinomial formula avoids error-prone sequential combination calculations.

3. Core mathematics

Multinomial coefficient

(n _{1} , n _{2} , \dots , n _{k} n) = \frac{n !}{n _{1} ! n _{2} ! \dots n _{k} !}

Special case k=2

(n _{1} , n _{2} n) = (n _{1} n)

Reduces to the ordinary binomial coefficient when there are only two groups.

4. How to recognize it

Think of this when…

Items are split into more than two labeled or distinguishable groups.
The question asks for distinct arrangements of a word or sequence with repeated symbols.
Phrases like "divided into teams of sizes..." with specific labeled team sizes.
You'd otherwise chain several combination calculations together — a multinomial coefficient does it in one step.

5. Worked example

Worked example

12 people are divided into three labeled teams: Team A with 5 people, Team B with 4 people, and Team C with 3 people. How many ways can this be done?

1
This is a direct application of the multinomial coefficient with $n = 12$ , $n_{1} = 5$ , $n_{2} = 4$ , $n_{3} = 3$ .
2
Write out the formula.
$(5 , 4 , 3 12) = \frac{12 !}{5 ! 4 ! 3 !}$
3
Compute: $12! = 479001600$ , $5! = 120$ , $4! = 24$ , $3! = 6$ , so denominator $= 120 \times 24 \times 6 = 17280$ .
$\frac{479001600}{17280} = 27720$
Answer: 27,720 ways.

6. Common mistakes

Dividing by an extra factor of $k!$ when the groups are labeled/distinguishable (that correction is only needed for unlabeled/interchangeable groups).
Forgetting that group sizes must sum to $n$ ; mismatched sizes signal an error in the setup.
Confusing multinomial coefficients (partitioning a set) with multinomial distribution probabilities (which also include a probability weighting term).
Using ordinary combinations sequentially but making an arithmetic slip in the remaining pool size at each step.

7. Quick check

How many distinct arrangements are there of the letters in "MISSISSIPPI" (11 letters: M=1, I=4, S=4, P=2)?

If instead the three teams in the worked example were unlabeled (indistinguishable) but still sized 5, 4, 3, would the count change?

8. Practice

12 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Multinomial Counting

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