Multinomial Theorem (Advanced)
Introduction
The Multinomial Theorem is a generalization of the Binomial Theorem. While binomial theorem deals with expansion of $(a + b)^n$, multinomial theorem expands $(x_1 + x_2 + x_3 + \cdots + x_k)^n$.
This is an advanced topic primarily for JEE Advanced and olympiad-level problems.
Multinomial Theorem Statement
Multinomial Theorem: For positive integer $n$ and any $k$ variables:
$$\boxed{(x_1 + x_2 + \cdots + x_k)^n = \sum \frac{n!}{n_1! n_2! \cdots n_k!} x_1^{n_1} x_2^{n_2} \cdots x_k^{n_k}}$$
where the sum is taken over all non-negative integers $n_1, n_2, \ldots, n_k$ such that:
$$n_1 + n_2 + \cdots + n_k = n$$Interactive Demo: Visualize Multinomial Patterns
Explore how multinomial coefficients extend Pascal’s Triangle.
Multinomial Coefficient
The coefficient $\frac{n!}{n_1! n_2! \cdots n_k!}$ is called the multinomial coefficient, denoted:
$$\boxed{\binom{n}{n_1, n_2, \ldots, n_k} = \frac{n!}{n_1! n_2! \cdots n_k!}}$$
Special Cases:
- When $k = 2$: $\binom{n}{n_1, n_2} = \binom{n}{n_1} = \frac{n!}{n_1! n_2!}$ (Binomial coefficient)
- When $k = 3$: $\binom{n}{n_1, n_2, n_3} = \frac{n!}{n_1! n_2! n_3!}$ (Trinomial coefficient)
Trinomial Expansion
The most common case after binomial is trinomial expansion:
$$\boxed{(a + b + c)^n = \sum_{p+q+r=n} \frac{n!}{p! q! r!} a^p b^q c^r}$$
where $p, q, r \geq 0$ and $p + q + r = n$.
Example: $(a + b + c)^2$
$$\begin{align} (a + b + c)^2 &= a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \end{align}$$Using multinomial theorem:
- Term $a^2$: $\frac{2!}{2!0!0!}a^2 = 1 \cdot a^2$
- Term $b^2$: $\frac{2!}{0!2!0!}b^2 = 1 \cdot b^2$
- Term $c^2$: $\frac{2!}{0!0!2!}c^2 = 1 \cdot c^2$
- Term $ab$: $\frac{2!}{1!1!0!}ab = 2ab$
- Term $bc$: $\frac{2!}{0!1!1!}bc = 2bc$
- Term $ca$: $\frac{2!}{1!0!1!}ca = 2ca$
Number of Terms
Formula for Number of Terms
In the expansion of $(x_1 + x_2 + \cdots + x_k)^n$:
$$\boxed{\text{Number of terms} = \binom{n + k - 1}{k - 1} = \binom{n + k - 1}{n}}$$
Derivation: This is equivalent to finding the number of non-negative integer solutions to:
$$n_1 + n_2 + \cdots + n_k = n$$which is a classic “stars and bars” problem.
Examples
$(a + b + c)^{10}$: Number of terms = $\binom{10+3-1}{3-1} = \binom{12}{2} = 66$
$(x + y + z + w)^5$: Number of terms = $\binom{5+4-1}{4-1} = \binom{8}{3} = 56$
General Term in Trinomial
For $(x + y + z)^n$, the general term is:
$$\boxed{T_{p,q,r} = \frac{n!}{p! q! r!} x^p y^q z^r}$$
where $p + q + r = n$ and $p, q, r \geq 0$.
Finding Specific Terms
Method 1: Direct Substitution
To find coefficient of $x^a y^b z^c$ in $(x + y + z)^n$:
- Check if $a + b + c = n$
- If yes, coefficient = $\frac{n!}{a! b! c!}$
- If no, coefficient = 0
Method 2: Grouping Method
To find coefficient in $(ax + by + cz)^n$:
Coefficient of $x^p y^q z^r$ = $\frac{n!}{p! q! r!} a^p b^q c^r$
Properties of Multinomial Coefficients
Property 1: Sum of All Coefficients
$$\boxed{\sum \binom{n}{n_1, n_2, \ldots, n_k} = k^n}$$
Proof: Put $x_1 = x_2 = \cdots = x_k = 1$ in multinomial theorem.
Example: Sum of all coefficients in $(a + b + c)^5 = 3^5 = 243$
Property 2: Multinomial Identity
$$\boxed{\sum_{n_1 + \cdots + n_k = n} \binom{n}{n_1, \ldots, n_k} = k^n}$$
Property 3: Relationship with Binomial
$$\binom{n}{n_1, n_2, \ldots, n_k} = \binom{n}{n_1} \binom{n - n_1}{n_2} \binom{n - n_1 - n_2}{n_3} \cdots$$Advanced Applications
Application 1: Greatest Coefficient
To find greatest multinomial coefficient in $(x_1 + \cdots + x_k)^n$:
The greatest coefficient occurs when the powers are as equal as possible.
If $n = kq + r$ where $0 \leq r < k$:
- $r$ terms get power $(q+1)$
- $(k-r)$ terms get power $q$
Application 2: Coefficient of Specific Power
Find coefficient of $x^m$ in $(1 + x + x^2 + \cdots + x^k)^n$:
This requires generating functions and is related to partition theory.
Memory Tricks
🎯 Multinomial Coefficient Pattern
“Factorial on top, factorials below”:
$$\frac{n!}{\text{product of factorials of individual powers}}$$🎯 Number of Terms
“Stars and Bars”:
- $n$ stars (representing the power)
- $k-1$ bars (separating into $k$ groups)
- Total arrangements: $\binom{n+k-1}{k-1}$
🎯 Sum of Coefficients
“Replace all variables with 1”: $(1 + 1 + \cdots + 1)^n = k^n$ where $k$ is number of terms.
Common Mistakes to Avoid
❌ Mistake 1: Wrong Factorial Arrangement
Wrong: Coefficient of $x^2y^3z^4$ in $(x+y+z)^9$ is $\frac{2!3!4!}{9!}$ ✗
Correct: Coefficient = $\frac{9!}{2!3!4!}$ ✓
Remember: Total factorial on top, individual factorials on bottom!
❌ Mistake 2: Forgetting Power Sum Condition
Wrong: Finding coefficient of $x^2y^3z^3$ in $(x+y+z)^7$ ✗
Correct: No such term exists since $2+3+3 = 8 \neq 7$ ✓
❌ Mistake 3: Incorrect Number of Terms Formula
Wrong: Number of terms in $(x+y+z)^n$ is $3n$ ✗
Correct: Number of terms = $\binom{n+2}{2}$ ✓
Solved Examples
Example 1: Trinomial Expansion (JEE Advanced)
Expand $(x + 2y - 3z)^3$ completely.
Solution: Using $(a + b + c)^3 = \sum_{p+q+r=3} \frac{3!}{p!q!r!} a^p b^q c^r$
Here $a = x$, $b = 2y$, $c = -3z$
Terms where $p+q+r=3$:
| $p$ | $q$ | $r$ | Coefficient | Term |
|---|---|---|---|---|
| 3 | 0 | 0 | $\frac{3!}{3!0!0!} = 1$ | $x^3$ |
| 0 | 3 | 0 | $\frac{3!}{0!3!0!} = 1$ | $(2y)^3 = 8y^3$ |
| 0 | 0 | 3 | $\frac{3!}{0!0!3!} = 1$ | $(-3z)^3 = -27z^3$ |
| 2 | 1 | 0 | $\frac{3!}{2!1!0!} = 3$ | $3x^2(2y) = 6x^2y$ |
| 2 | 0 | 1 | $\frac{3!}{2!0!1!} = 3$ | $3x^2(-3z) = -9x^2z$ |
| 1 | 2 | 0 | $\frac{3!}{1!2!0!} = 3$ | $3x(2y)^2 = 12xy^2$ |
| 0 | 2 | 1 | $\frac{3!}{0!2!1!} = 3$ | $3(2y)^2(-3z) = -36y^2z$ |
| 1 | 0 | 2 | $\frac{3!}{1!0!2!} = 3$ | $3x(-3z)^2 = 27xz^2$ |
| 0 | 1 | 2 | $\frac{3!}{0!1!2!} = 3$ | $3(2y)(-3z)^2 = 54yz^2$ |
| 1 | 1 | 1 | $\frac{3!}{1!1!1!} = 6$ | $6x(2y)(-3z) = -36xyz$ |
Final Answer:
$$x^3 + 8y^3 - 27z^3 + 6x^2y - 9x^2z + 12xy^2 - 36y^2z + 27xz^2 + 54yz^2 - 36xyz$$Example 2: Finding Coefficient (JEE Advanced)
Find the coefficient of $x^3y^4z^2$ in the expansion of $(x + y + z)^9$.
Solution: Check: $3 + 4 + 2 = 9$ ✓
Coefficient = $\frac{9!}{3! \cdot 4! \cdot 2!}$
$$= \frac{362880}{6 \times 24 \times 2} = \frac{362880}{288} = 1260$$Example 3: Number of Terms (JEE Advanced)
How many distinct terms are there in the expansion of $(a + b + c + d)^{10}$?
Solution: Using formula: Number of terms = $\binom{n+k-1}{k-1}$
Here $n = 10$, $k = 4$:
$$\text{Number of terms} = \binom{10+4-1}{4-1} = \binom{13}{3} = \frac{13 \times 12 \times 11}{6} = 286$$Example 4: Sum of Coefficients (JEE Advanced)
Find the sum of all coefficients in $(2x - 3y + 4z)^{15}$.
Solution: Put $x = y = z = 1$:
Sum = $(2(1) - 3(1) + 4(1))^{15} = (2 - 3 + 4)^{15} = 3^{15}$
Example 5: Coefficient with Powers (JEE Advanced)
Find the coefficient of $x^2y^3z^4$ in $(2x - y + 3z)^9$.
Solution: Check: $2 + 3 + 4 = 9$ ✓
General term: $\frac{9!}{p!q!r!}(2x)^p(-y)^q(3z)^r$ where $p+q+r=9$
For $x^2y^3z^4$: $p=2, q=3, r=4$
Coefficient = $\frac{9!}{2!3!4!} \times 2^2 \times (-1)^3 \times 3^4$
$$= \frac{362880}{2 \times 6 \times 24} \times 4 \times (-1) \times 81$$ $$= 1260 \times 4 \times (-1) \times 81 = -408240$$Practice Problems
Level 1: JEE Main/Advanced Basics
Problem 1.1: Find the number of terms in $(a + b + c)^8$.
Solution
Number of terms = $\binom{8+3-1}{3-1} = \binom{10}{2} = 45$
Problem 1.2: Find the coefficient of $abc$ in $(a + b + c)^3$.
Solution
For $abc$: powers are $(1,1,1)$ and sum = 3 ✓
Coefficient = $\frac{3!}{1!1!1!} = 6$
Problem 1.3: Find sum of all coefficients in $(x + y + z)^{10}$.
Solution
Put $x = y = z = 1$: Sum = $(1+1+1)^{10} = 3^{10} = 59049$
Level 2: JEE Advanced
Problem 2.1: Find the coefficient of $x^2y^2z^2$ in $(x + y + z)^6$.
Solution
Check: $2 + 2 + 2 = 6$ ✓
Coefficient = $\frac{6!}{2!2!2!} = \frac{720}{8} = 90$
Problem 2.2: Find the greatest coefficient in $(1 + x + x^2)^{10}$.
Solution
Rewrite as $(1 + x + x^2)^{10} = \sum \frac{10!}{p!q!r!} 1^p x^q (x^2)^r$
$= \sum \frac{10!}{p!q!r!} x^{q+2r}$ where $p+q+r=10$
Greatest multinomial coefficient when powers are most equal: $p = q = r = 3$ with one being 4 is not possible since we need $p+q+r=10$.
Try: $p=4, q=3, r=3$: Coefficient = $\frac{10!}{4!3!3!} = 4200$
Or $p=3, q=4, r=3$: Same = $4200$
Or $p=3, q=3, r=4$: Same = $4200$
Greatest coefficient = $4200$
Problem 2.3: In $(1 + x + x^2 + x^3)^{100}$, find the coefficient of $x^{200}$.
Solution
$(1 + x + x^2 + x^3)^{100} = \sum \frac{100!}{n_0!n_1!n_2!n_3!} 1^{n_0} x^{n_1} (x^2)^{n_2} (x^3)^{n_3}$
where $n_0 + n_1 + n_2 + n_3 = 100$ and power of $x = n_1 + 2n_2 + 3n_3 = 200$
This is a complex partition problem. We need to count solutions to:
- $n_0 + n_1 + n_2 + n_3 = 100$
- $n_1 + 2n_2 + 3n_3 = 200$
From second: $n_1 = 200 - 2n_2 - 3n_3$
Substitute in first: $n_0 + 200 - 2n_2 - 3n_3 + n_2 + n_3 = 100$
$n_0 = -100 + n_2 + 2n_3$
Since $n_0 \geq 0$: $n_2 + 2n_3 \geq 100$
Also $n_1 \geq 0$: $2n_2 + 3n_3 \leq 200$
This requires systematic enumeration. (Advanced problem)
Level 3: Olympiad Level
Problem 3.1: Prove that the sum of all multinomial coefficients $\binom{n}{k_1, k_2, \ldots, k_m}$ where $k_1 + k_2 + \cdots + k_m = n$ equals $m^n$.
Solution
Consider $(x_1 + x_2 + \cdots + x_m)^n = \sum \binom{n}{k_1,\ldots,k_m} x_1^{k_1} \cdots x_m^{k_m}$
Put $x_1 = x_2 = \cdots = x_m = 1$:
$(1 + 1 + \cdots + 1)^n = m^n = \sum \binom{n}{k_1,\ldots,k_m}$
Problem 3.2: Find the coefficient of $x^{50}$ in $(1 + x + x^2 + \cdots + x^{10})^{10}$.
Solution
Using generating functions:
$(1 + x + x^2 + \cdots + x^{10})^{10} = \left(\frac{1-x^{11}}{1-x}\right)^{10}$
$= (1-x^{11})^{10}(1-x)^{-10}$
$= \left[\sum_{k=0}^{10}\binom{10}{k}(-x^{11})^k\right] \left[\sum_{j=0}^{\infty}\binom{-10}{j}(-x)^j\right]$
$= \left[\sum_{k=0}^{10}\binom{10}{k}(-1)^k x^{11k}\right] \left[\sum_{j=0}^{\infty}\binom{j+9}{9}x^j\right]$
Coefficient of $x^{50}$: Need $11k + j = 50$
- $k=0, j=50$: $\binom{10}{0} \times 1 \times \binom{59}{9}$
- $k=1, j=39$: $\binom{10}{1} \times (-1) \times \binom{48}{9}$
- $k=2, j=28$: $\binom{10}{2} \times 1 \times \binom{37}{9}$
- $k=3, j=17$: $\binom{10}{3} \times (-1) \times \binom{26}{9}$
- $k=4, j=6$: $\binom{10}{4} \times 1 \times \binom{15}{9}$
Sum these to get the coefficient.
Important Formulas Summary
| Property | Formula |
|---|---|
| Multinomial Expansion | $(x_1 + \cdots + x_k)^n = \sum \frac{n!}{n_1!\cdots n_k!}x_1^{n_1}\cdots x_k^{n_k}$ |
| Multinomial Coefficient | $\binom{n}{n_1,\ldots,n_k} = \frac{n!}{n_1! \cdots n_k!}$ |
| Number of Terms | $\binom{n+k-1}{k-1}$ |
| Sum of Coefficients | $k^n$ (put all variables = 1) |
| Trinomial | $(a+b+c)^n = \sum_{p+q+r=n}\frac{n!}{p!q!r!}a^pb^qc^r$ |
Cross-References
- Binomial Theorem: Special case of multinomial → Binomial Expansion
- Combinatorics: Stars and bars method → Combinatorics
- Generating Functions: Advanced applications → Sequences and Series
Quick Revision Points
- Multinomial generalizes binomial to multiple terms
- Coefficient: Total factorial divided by individual factorials
- Number of terms: Use stars and bars formula
- Sum of coefficients: Replace all variables with 1
- Always check if power sum equals $n$
Last updated: October 28, 2025