Multinomial Expansion
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Direct answer
Two variables ask "which r"; three or more ask "which whole composition of n", and the multinomial theorem answers with n!/(p!q!r!...) instead of a single C(n, r). Precisely: (x₁ + x₂ + ... + x_k)^n = Σ [n!/(a₁!a₂!...a_k!)]·x₁^a₁...x_k^a_k, summed over all non-negative compositions a₁ + ... + a_k = n. Each coefficient counts the ways to decide which of the n factors supplies which variable — an ordered partition of n objects into k labelled groups. The number of distinct terms is C(n + k − 1, k − 1), the same stars-and-bars count that solves integer-solution equations.
What you must remember
- The theorem: (x + y + z)^n = Σ [n!/(p!q!r!)]·x^p y^q z^r over p + q + r = n; the coefficient of x^p y^q z^r is n!/(p!q!r!) when p + q + r = n and 0 otherwise.
- Term count: (x₁ + ... + x_k)^n has C(n + k − 1, k − 1) distinct terms; (a + b + c)^5 has C(7, 2) = 21.
- Binomial collapse: fixing p and summing over the rest recovers C(n, p), since n!/(p!q!r!) = C(n, p)·(n − p)!/(q!r!) — the multinomial theorem contains the binomial one.
- Coefficient extraction: the coefficient of x² in (1 + x + x²)^5 is 15 — five ways to take x² from one factor plus C(5, 2) = 10 ways to take x from two factors.
- Total coefficient sum: set every variable equal to 1 and the expansion collapses to k^n, the sum of all multinomial coefficients — the count of functions from an n-set to a k-set.
- Counting link: the number of non-negative integer solutions of a₁ + ... + a_k = n is C(n + k − 1, k − 1) — one formula serving two chapters.
- Zero exponents allowed: terms like x^n or x^p y^q (r = 0) are legitimate, and the factorial formula holds unchanged; compositions need only be non-negative, not positive.
Coefficient of a three-variable term
Find the coefficient of x³y²z² in (x + 2y + z)^7. Step one: solve the composition 3 + 2 + 2 = 7 — unique, so the term exists. Step two: apply the factorial formula 7!/(3!·2!·2!) = 5040/24 = 210. Step three: attach the constants, here 2² from the coefficient 2 riding each of the two y-factors, giving 210 × 4 = 840. Three steps, no expansion of a 7th power — and the same routine works when the constant part participates, as in (x + y + 1)^6: the coefficient of x³y² there splits over how much degree the 1 absorbs, giving C(6,3)C(3,2) + C(6,3)C(3,2)·1/1? — concretely, compositions (3, 2, 1) and (3, 2, 0) both contribute: 6!/(3!2!1!) + 6!/(3!2!0!) = 60 + 60 = 120.
Verify small claims with the substitution trick: putting every variable equal to 1 in (1 + x + x²)^5 must give 3⁵ = 243, and summing all coefficients — including the 15 above — does total 243. This one-line global check catches most coefficient-extraction slips before the answer is committed.
Where the term count trips you
Main rarely sells the multinomial theorem openly; it hides inside permutations-and-combinations questions (distributing n distinct objects into k labelled boxes) and coefficient extractions, while Advanced uses it for multiple-correct coefficient comparisons across several expansions. The recurring confusions: mixing up the term count C(n + k − 1, k − 1) with the coefficient total k^n — one counts distinct monomials, the other counts the sum of all coefficients; forgetting that zero exponents are permitted, so x⁵ is a genuine term of (x + y + z)^5; and mishandling bounded compositions, where each factor of (1 + x + x²)^n can contribute at most degree 2, making the coefficient a count of restricted compositions rather than a plain factorial ratio.
Frequently asked questions
How many terms does (a + b + c)^n contain?
C(n + 2, 2) distinct terms — the number of non-negative solutions of p + q + r = n; for n = 5 that is 21.
What is the coefficient of x^p y^q z^r in (x + y + z)^n?
n!/(p!q!r!) whenever p + q + r = n, and 0 otherwise.
How is the multinomial theorem linked to combinations?
Each coefficient counts the ways to choose which of the n identical factors supplies x, y and z — an ordered partition of n positions into labelled groups.
Why does substituting 1 for every variable give k^n?
The expansion then lists every way each of n factors picks one of k variables, so all coefficients together must sum to k^n.
Can exponents in a multinomial term be zero?
Yes — terms like x^n or x^p y^q with r = 0 are valid, and the coefficient formula n!/(p!q!r!) applies unchanged.