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The multinomial coefficient n!/(k₁! k₂! ⋯ kₘ!) counts the number of ways to divide n distinct objects into m ordered groups of sizes k₁, k₂, …, kₘ where k₁ + k₂ + ⋯ + kₘ = n. The multinomial theorem generalizes the binomial theorem: (x₁ + x₂ + ⋯ + xₘ)ⁿ equals the sum over all valid (k₁,…,kₘ) of the multinomial coefficient times x₁^k₁ ⋯ xₘ^kₘ. Multinomial coefficients arise naturally when counting arrangements of strings with repeated letters.
Connect to the binomial theorem first as the m=2 special case. Count arrangements of words with repeated letters (e.g., MISSISSIPPI has 11!/(4!4!2!1!) arrangements) to make the formula concrete before moving to polynomial expansion.
You already know the binomial coefficient C(n, r) = n!/(r!(n−r)!) and the binomial theorem: (x + y)ⁿ expands into a sum of terms C(n, k) xᵏ yⁿ⁻ᵏ. The multinomial coefficient generalizes both to situations with more than two categories. Suppose you have n distinct objects and want to sort them into m labeled boxes of sizes k₁, k₂, …, kₘ (where the sizes sum to n). The multinomial coefficient n!/(k₁! k₂! ⋯ kₘ!) counts the ways to do this. When m = 2, it reduces exactly to C(n, k₁) = n!/(k₁! k₂!) — the familiar binomial coefficient.
The most concrete way to build intuition is through arrangements of strings with repeated letters. How many ways can you arrange the letters in MISSISSIPPI? You have 11 letters: 1 M, 4 I's, 4 S's, and 2 P's. If all 11 were distinct, there'd be 11! arrangements. But the 4 I's are identical — any permutation among them gives the same word — so divide by 4!. Same for the S's (divide by 4!) and P's (divide by 2!). Result: 11!/(1!⋅4!⋅4!⋅2!) = 34,650. The formula is not an arbitrary definition; it's exactly the correction factor for repeated elements.
The multinomial theorem extends this to polynomial expansion: (x₁ + x₂ + ⋯ + xₘ)ⁿ = Σ n!/(k₁!⋯kₘ!) · x₁^k₁ ⋯ xₘ^kₘ, where the sum runs over all tuples (k₁,…,kₘ) of non-negative integers summing to n. To see why: expanding the product means choosing, from each of the n factors, one variable xᵢ to "take." For a given monomial x₁^k₁⋯xₘ^kₘ to appear, you must pick x₁ exactly k₁ times, x₂ exactly k₂ times, etc. — and the number of ways to make those selections is precisely the multinomial coefficient. The binomial theorem is just the m = 2 special case.
Where multinomial coefficients diverge from iterated binomial coefficients is subtle. You *can* compute a multinomial coefficient as a product of binomials: n!/(k₁!k₂!k₃!) = C(n, k₁) · C(n−k₁, k₂) · C(n−k₁−k₂, k₃). This "sequential selection" decomposition is valid and sometimes useful, but the end result — the multinomial coefficient — is a single number, not a sequence-dependent product. The value is independent of which decomposition order you choose, because the formula n!/(k₁!⋯kₘ!) is symmetric in interpretation.