In short

The multinomial theorem expands (x_1 + x_2 + \cdots + x_k)^n as a sum of terms \frac{n!}{n_1!\,n_2!\,\cdots\,n_k!}\,x_1^{n_1}x_2^{n_2}\cdots x_k^{n_k}, where the sum runs over all non-negative integer tuples (n_1, n_2, \ldots, n_k) with n_1 + n_2 + \cdots + n_k = n. Each coefficient \binom{n}{n_1, n_2, \ldots, n_k} counts the number of ways to assign n factors to k variable slots with the given sizes.

A sweet shop in your town sells three flavours of barfi — kaju, pista, and badam. You place an order for 4 pieces, and for each piece you independently pick one of the three flavours. If you wanted to count how many orders have exactly 2 kaju, 1 pista, and 1 badam, you could list them by hand. But what if the order had 10 pieces and 5 flavours? The multinomial theorem gives you a single formula that handles every such count at once, by connecting it to the expansion of (x_1 + x_2 + \cdots + x_k)^n.

From the binomial to the multinomial

You already know the binomial theorem: expanding (a + b)^n gives

(a + b)^n = \sum_{r=0}^{n} \binom{n}{r}\,a^{n-r}\,b^r

Each term corresponds to choosing r of the n factors to contribute b (and the remaining n - r contribute a). The coefficient \binom{n}{r} = \frac{n!}{r!(n-r)!} counts those choices.

Now replace two variables with three: expand (x + y + z)^n. Each of the n factors is (x + y + z), and from each factor you pick exactly one of x, y, or z. A typical term looks like x^a y^b z^c where a + b + c = n — you picked x from a factors, y from b factors, and z from c factors. How many ways can that happen? You are distributing n labelled factors into three groups of sizes a, b, and c. That count is the multinomial coefficient:

\frac{n!}{a!\,b!\,c!}
From binomial to multinomial: each factor contributes one variableFour boxes labelled Factor 1 through Factor 4 represent the expansion of (x+y+z) to the fourth power. Each box has three outgoing arrows labelled x, y, z. One specific choice — x from factors 1 and 3, y from factor 2, z from factor 4 — is highlighted, producing the term x squared y z. (x + y + z)⁴ — each factor contributes one variable Factor 1 Factor 2 Factor 3 Factor 4 x y x z This choice → x · y · x · z = x²yz Coefficient of x²yz counts how many choices give this term: 4!/(2!·1!·1!) = 12
Expanding $(x+y+z)^4$: each of the $4$ factors contributes $x$, $y$, or $z$. One specific choice produces the term $x^2yz$. The coefficient of $x^2yz$ is $\frac{4!}{2!\cdot1!\cdot1!} = 12$, counting all ways to assign the $4$ factors to produce exactly two $x$'s, one $y$, and one $z$.

The pattern extends to any number of variables. Expanding (x_1 + x_2 + \cdots + x_k)^n, each of the n factors contributes one of the k variables, and the coefficient of x_1^{n_1}x_2^{n_2}\cdots x_k^{n_k} counts the number of ways to assign the n factors so that exactly n_i of them contribute x_i.

The multinomial theorem — formal statement

Multinomial Theorem

For any positive integer n and variables x_1, x_2, \ldots, x_k:

(x_1 + x_2 + \cdots + x_k)^n = \sum_{\substack{n_1 + n_2 + \cdots + n_k = n \\ n_i \geq 0}} \frac{n!}{n_1!\,n_2!\,\cdots\,n_k!}\;x_1^{n_1}\,x_2^{n_2}\,\cdots\,x_k^{n_k}

The coefficient \frac{n!}{n_1!\,n_2!\,\cdots\,n_k!} is called the multinomial coefficient, written \binom{n}{n_1, n_2, \ldots, n_k}.

Why the formula is true. Multiplying out (x_1 + x_2 + \cdots + x_k)^n means choosing one term from each of the n copies of (x_1 + x_2 + \cdots + x_k) and multiplying the choices together. A generic product has the form x_1^{n_1}x_2^{n_2}\cdots x_k^{n_k} where n_1 + n_2 + \cdots + n_k = n. The number of ways to get this particular product is the number of ways to designate which n_1 factors contribute x_1, which n_2 contribute x_2, and so on. That is the multinomial coefficient — the number of ways to divide n labelled objects into k groups of sizes n_1, n_2, \ldots, n_k.

Structure of the multinomial expansionA diagram showing the multinomial expansion as a sum over all tuples (n1, n2, ..., nk) that sum to n. Each tuple contributes a term with a multinomial coefficient times a product of powers. The total number of such tuples is C(n+k-1, k-1). Structure of the expansion Pick exponents: n₁+n₂+…+nₖ = n → Coefficient: n! / (n₁!·n₂!·…·nₖ!) Each term: [coefficient] × x₁ⁿ¹ x₂ⁿ² … xₖⁿᵏ Sum runs over all non-negative integer solutions of n₁+n₂+…+nₖ = n Number of terms = C(n+k−1, k−1)
Each term in the multinomial expansion corresponds to one non-negative integer solution of $n_1 + n_2 + \cdots + n_k = n$. The coefficient of that term is the multinomial coefficient $\frac{n!}{n_1! n_2! \cdots n_k!}$, and the total number of distinct terms is $\binom{n+k-1}{k-1}$.

Checking against the binomial theorem. With k = 2, write x_1 = a and x_2 = b. The multinomial coefficient becomes \frac{n!}{r!(n-r)!} = \binom{n}{r}, and the sum runs over r = 0, 1, \ldots, n. The expansion reproduces (a+b)^n = \sum_{r=0}^n \binom{n}{r}a^{n-r}b^r exactly. The multinomial theorem is the natural generalisation.

Number of terms in the expansion

Each term corresponds to a non-negative integer solution of n_1 + n_2 + \cdots + n_k = n. By the stars and bars formula, the number of such solutions is:

\binom{n + k - 1}{k - 1}

For (x + y + z)^4 (where n = 4, k = 3): \binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = 15 distinct terms. For (a + b)^4 (where k = 2): \binom{5}{1} = 5 terms, matching the familiar binomial expansion a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4.

Number of terms in the expansion for various n and kA table showing the number of terms C(n+k-1, k-1) for n from 2 to 5 and k from 2 to 4. For n=4 k=3, the value 15 is highlighted. Number of terms = C(n+k−1, k−1) n \ k k = 2 k = 3 k = 4 n = 2 3 6 10 n = 3 4 10 20 n = 4 5 15 35 n = 5 6 21 56 (x+y+z)⁴ has C(6,2) = 15 terms (highlighted)
The number of terms in $(x_1 + \cdots + x_k)^n$ grows quickly with $k$ and $n$. The highlighted cell confirms $(x+y+z)^4$ has $\binom{6}{2} = 15$ terms.

The sum of all multinomial coefficients

Setting x_1 = x_2 = \cdots = x_k = 1 in the expansion gives:

(1 + 1 + \cdots + 1)^n = k^n = \sum_{\substack{n_1 + \cdots + n_k = n}} \frac{n!}{n_1!\,n_2!\,\cdots\,n_k!}

The sum of all multinomial coefficients in the expansion of (x_1 + x_2 + \cdots + x_k)^n equals k^n. For the binomial case (k = 2), this gives the familiar identity \sum_{r=0}^n \binom{n}{r} = 2^n.

This identity has a direct combinatorial reading: k^n is the number of ways to assign each of n labelled factors to one of k variables. Every such assignment contributes to exactly one multinomial coefficient, and the total over all coefficients is k^n.

Applications

The multinomial theorem is not just an algebraic expansion — it counts things.

Finding a specific coefficient. To find the coefficient of x^2y^3z in (x + y + z)^6, identify the exponents: n_1 = 2, n_2 = 3, n_3 = 1, with n = 6. The coefficient is:

\frac{6!}{2!\,3!\,1!} = \frac{720}{2 \times 6 \times 1} = 60

Counting rearrangements of letters. The number of distinct ways to rearrange the letters of the word MISSISSIPPI (11 letters: 1 M, 4 I, 4 S, 2 P) is the multinomial coefficient:

\frac{11!}{1!\,4!\,4!\,2!} = \frac{39{,}916{,}800}{1 \times 24 \times 24 \times 2} = 34{,}650

This is not a coincidence. Rearranging n objects where there are n_1 copies of type 1, n_2 copies of type 2, and so on, is exactly the same as distributing n positions among k types — and that count is the multinomial coefficient.

Rearrangements of MISSISSIPPI as a multinomial coefficientThe word MISSISSIPPI is shown with each letter type colour-coded. Below, the formula 11 factorial divided by 1 factorial times 4 factorial times 4 factorial times 2 factorial equals 34650 is displayed. MISSISSIPPI — 11 letters M I S S I S S I P P I 1M, 4I, 4S, 2P Distinct arrangements = 11! / (1!·4!·4!·2!) = 39,916,800 / 1,152 = 34,650 Each arrangement is one term in the multinomial expansion of (M+I+S+P)¹¹
The letters of MISSISSIPPI: $1$ M, $4$ I's, $4$ S's, $2$ P's. The number of distinct rearrangements is the multinomial coefficient $\frac{11!}{1!\cdot 4!\cdot 4!\cdot 2!} = 34{,}650$.

Two worked examples

Example 1: Find the coefficient of $x^2 y^2 z$ in the expansion of $(x + y + z)^5$.

Step 1. Identify the exponents.

The term x^2 y^2 z has exponents n_1 = 2, n_2 = 2, n_3 = 1. Check: 2 + 2 + 1 = 5 = n. The exponents sum to the power, so this term does appear in the expansion.

Why: a term x_1^{n_1}\cdots x_k^{n_k} appears in (x_1 + \cdots + x_k)^n only when n_1 + \cdots + n_k = n. If the exponents do not sum to n, the term is absent.

Step 2. Apply the multinomial coefficient formula.

\frac{n!}{n_1!\,n_2!\,n_3!} = \frac{5!}{2!\,2!\,1!} = \frac{120}{2 \times 2 \times 1} = 30

Why: this counts the number of ways to choose which 2 of the 5 factors contribute x, which 2 contribute y, and which 1 contributes z.

Step 3. Verify by breaking into binomial steps.

Think of the 5 factors. Choose 2 for x: \binom{5}{2} = 10 ways. From the remaining 3, choose 2 for y: \binom{3}{2} = 3 ways. The last factor must contribute z: \binom{1}{1} = 1 way. Total: 10 \times 3 \times 1 = 30.

Why: this sequential selection method matches the multinomial coefficient. The product \binom{5}{2}\binom{3}{2}\binom{1}{1} telescopes to \frac{5!}{2!\,2!\,1!} after cancellation.

Step 4. Sanity check: sum of all coefficients.

The total of all coefficients in (x+y+z)^5 is 3^5 = 243. The coefficient 30 is roughly 30/243 \approx 12\% of the total, which is reasonable for one of the \binom{7}{2} = 21 terms.

Result: The coefficient of x^2 y^2 z in (x + y + z)^5 is \mathbf{30}.

Coefficient of x squared y squared z in (x+y+z) to the fifthFive boxes represent the five factors. Two are assigned to x (highlighted red), two to y, one to z. The sequential counting is shown: C(5,2) times C(3,2) times C(1,1) equals 10 times 3 times 1 equals 30. 5 factors: assign 2→x, 2→y, 1→z F1 → x F2 → x F3 → y F4 → y F5 → z Choose 2 for x: C(5, 2) = 10 Choose 2 for y: C(3, 2) = 3 Remaining for z: C(1, 1) = 1 Total = 10 × 3 × 1 = 30
One specific assignment of the $5$ factors: factors $1$ and $2$ contribute $x$, factors $3$ and $4$ contribute $y$, factor $5$ contributes $z$. The sequential counting $\binom{5}{2} \times \binom{3}{2} \times \binom{1}{1} = 30$ matches the multinomial coefficient $\frac{5!}{2!\,2!\,1!} = 30$.

The diagram makes the logic visible: the 30 different assignments of factors to variables produce 30 copies of the term x^2y^2z in the expanded product.

Example 2: Find the coefficient of $a^3 b$ in the expansion of $(2a - b)^4$.

Step 1. Rewrite to match the multinomial form.

Write (2a - b)^4 = (2a + (-b))^4. This is a binomial expansion (a special case of the multinomial with k = 2), where x_1 = 2a and x_2 = -b.

Why: the multinomial theorem applies to sums. The negative sign is absorbed into the variable: -b is just another term being added.

Step 2. Identify the exponents and apply the formula.

The term a^3 b requires x_1^3 x_2^1 = (2a)^3(-b)^1, so n_1 = 3 and n_2 = 1. The multinomial (here, binomial) coefficient is:

\binom{4}{3, 1} = \frac{4!}{3!\,1!} = 4

Why: with n = 4, n_1 = 3, n_2 = 1, the coefficient counts the ways to choose which 3 of the 4 factors contribute 2a and which 1 contributes -b.

Step 3. Include the constants from x_1 = 2a and x_2 = -b.

(2a)^3(-b)^1 = 8a^3 \times (-b) = -8a^3 b

The full term is: 4 \times (-8a^3b) = -32a^3b.

Why: the multinomial coefficient (4) counts the assignments, and the actual term value includes the constants (2^3 = 8) and the sign ((-1)^1 = -1) from the original expression.

Step 4. Verify using direct expansion.

(2a - b)^4 = (2a)^4 - 4(2a)^3(b) + 6(2a)^2(b)^2 - 4(2a)(b)^3 + b^4 = 16a^4 - 32a^3b + 24a^2b^2 - 8ab^3 + b^4

The coefficient of a^3b is -32. This matches.

Result: The coefficient of a^3b in (2a - b)^4 is \mathbf{-32}.

Coefficient of a cubed b in (2a minus b) to the fourthA flow diagram. Top: 4 factors of (2a + (-b)). Three factors contribute 2a and one contributes negative b. The multinomial coefficient 4 is multiplied by (2a) cubed times (-b) to get negative 32 a cubed b. (2a − b)⁴: coefficient of a³b 3 factors → 2a 1 factor → (−b) C(4; 3,1) = 4!/3!1! = 4 4 × (2a)³ × (−b)¹ = 4 × 8 × (−1) × a³b = −32 a³b
The coefficient of $a^3b$ in $(2a - b)^4$. The multinomial coefficient $\binom{4}{3,1} = 4$ is multiplied by $(2a)^3(-b)^1 = -8a^3b$, giving $-32a^3b$. The negative sign comes from the $(-b)^1$ factor.

The sign matters. When the original expression contains subtraction, the multinomial coefficient is always positive, but the term itself can be positive or negative depending on the power of the negative part: (-b)^{n_2} contributes (-1)^{n_2}.

Common confusions

Going deeper

If you can identify exponents, compute the multinomial coefficient, handle numerical coefficients and signs, and count the number of terms, you have what you need for exam-level problems. The rest of this section connects the multinomial theorem to broader ideas in algebra and counting.

The multinomial as a product of binomials

The multinomial coefficient satisfies a telescoping product:

\binom{n}{n_1, n_2, \ldots, n_k} = \binom{n}{n_1} \binom{n - n_1}{n_2} \binom{n - n_1 - n_2}{n_3} \cdots \binom{n_k}{n_k}

Each factor in the product is a standard binomial coefficient. This identity reflects the sequential process: first choose n_1 positions for x_1, then n_2 from the remainder for x_2, and so on. The last factor is always \binom{n_k}{n_k} = 1.

The general term of (a_1 x_1 + a_2 x_2 + \cdots + a_k x_k)^n

When the variables carry numerical coefficients a_1, a_2, \ldots, a_k, the general term becomes:

\frac{n!}{n_1!\,n_2!\,\cdots\,n_k!}\;a_1^{n_1}\,a_2^{n_2}\,\cdots\,a_k^{n_k}\;x_1^{n_1}\,x_2^{n_2}\,\cdots\,x_k^{n_k}

The multinomial coefficient is unchanged; the extra a_i^{n_i} factors come from raising each a_i x_i to the power n_i. This is exactly what happened in Example 2 with (2a - b)^4, where a_1 = 2 and a_2 = -1.

Connection to the number of surjections

The multinomial theorem can be used to count the number of surjections (onto functions) from a set of n elements to a set of k elements. By inclusion-exclusion, the number of surjections is:

\sum_{j=0}^{k} (-1)^j \binom{k}{j}(k - j)^n

The factor (k-j)^n is itself a sum of multinomial coefficients (via (x_1 + \cdots + x_{k-j})^n evaluated at x_i = 1). This connection to the inclusion-exclusion principle runs deep: multinomial expansions and inclusion-exclusion are two sides of the same combinatorial coin.

Where this leads next