Multinomial Coefficients
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Direct answer
When an expansion involves three or more letters, the multinomial theorem gives the general term of (x₁ + x₂ + ... + x_k)^n as [n!/(n₁!·n₂!...n_k!)]·x₁^n₁·x₂^n₂...x_k^n_k with n₁ + n₂ + ... + n_k = n. The same coefficient counts the distinct arrangements of a word with repeated letters — MATHEMATICS has 11 letters with T, A, M each appearing twice, giving 11!/(2!·2!·2!) = 4989600 — and the number of ways to distribute n distinct objects into labelled groups of sizes n₁, ..., n_k. The binomial theorem is the k = 2 case, so C(n, r) = n!/(r!(n − r)!) is literally a multinomial coefficient with two parts. Every multinomial question is either "find the coefficient" or "count the arrangements"; both resolve through the same factorial ratio.
What you must remember
- General term: for (x + y + z)^n, the term containing x^a·y^b·z^c has coefficient n!/(a!·b!·c!) with a + b + c = n; the number of distinct terms is C(n + 2, 2) for three letters (C(n + k − 1, k − 1) in general).
- Sum of all multinomial coefficients: (1 + 1 + ... + 1)^n = k^n — the identity that links expansions to distribution counts.
- Repeated-letter words: arrangements = (total letters)!/(product of factorials of repeat counts); BANANA = 6!/(3!·2!) = 60.
- Distribution link: n distinct objects into labelled boxes with fixed occupancies n₁, ..., n_k: n!/(n₁!...n_k!) — same ratio, different story.
- Greatest multinomial coefficient: for (x₁ + ... + x_k)^n, maxima occur when the nᵢ are as equal as possible (differing by at most 1); for n divisible by k, the peak is n!/((n/k)!^k).
- Coefficient extraction with powers: in (a + b + c)^10, the coefficient of a⁴b³c³ is 10!/(4!·3!·3!) = 4200 — arithmetic that should take seconds.
- Composite expansions: expand (1 + x + x²)^n by treating it as a geometric sub-series or by the multinomial with three parts; both must agree on, say, the coefficient of x² (which is C(n,1) + C(n,2)).
One coefficient, three letters
Find the coefficient of a⁴b³c³ in (a + b + c)^10. The exponents sum to 10, so the multinomial applies directly: 10!/(4!·3!·3!) = 3628800/864 = 4200. Now the same machinery on words: how many distinct strings can be formed from MISSISSIPPI? Four S, four I, two P, one M: 11!/(4!·4!·2!) = 34650. And on distribution: 9 distinct prizes into piles of 4, 3, 2 for three (unlabelled-by-content but named) children: 9!/(4!·3!·2!) = 1260. Three questions, one factorial ratio — the examiners' favourite economy, and the reason this page is quick marks once the pattern is visible.
The technique for composite bases deserves one walkthrough: the coefficient of x³ in (1 + x − 2x²)^4. Each factor contributes 1, x, or −2x²; to reach x³ either three x's (C(4,3)·1³ = 4) or one x and one x² (choose the x² factor 4 ways, the x factor 3 ways: 12 terms × 1 × (−2) = −24). Total: 4 − 24 = −20. This case-by-case assembly is the multinomial theorem used as a thinking tool rather than a formula drop.
Why multinomial questions are quick marks
JEE Main extracts coefficients with exponents that sum correctly, and the entire difficulty is discipline: verify the exponents add to n before dividing — a mismatched triple like a⁴b³c⁴ in degree 10 signals either a typo in reading or a need to first apply binomial on a grouped base. Advanced prefers the composite-base type and word problems with a twist: "words from EQUATION with at least two vowels together" first counts arrangements of the repeated blocks. Two recurring traps: forgetting that the number of terms in (a + b + c)^n is C(n + 2, 2), not 3^n (that counts all strings including positional identity); and mixing up k^n (functions from n objects to k labelled boxes) with n!/(n₁!...) (fixed occupancies). Keep the two stories separate and the factorial ratio never lies.
Frequently asked questions
What is the general term in the expansion of (a + b + c)^n?
[n!/(p!·q!·r!)]·a^p·b^q·c^r with p + q + r = n.
How many distinct arrangements does MISSISSIPPI have?
11!/(4!·4!·2!) = 34650, dividing by the factorials of each repeated letter's count.
What is the coefficient of a⁴b³c³ in (a + b + c)^10?
10!/(4!·3!·3!) = 4200, since the exponents sum to the power 10.
What is the sum of all multinomial coefficients in (x₁ + ... + x_k)^n?
k^n, by substituting every variable equal to 1.
How many terms does (a + b + c)^n contain?
C(n + 2, 2) = (n + 1)(n + 2)/2 distinct terms — combinations with repetition of the exponent triple, not 3^n.