The Multinomial Theorem: A Bridge to Binomial & Beyond
Master the generalization of binomial expansion to multiple variables with applications in JEE Advanced problems.
Why Multinomial Theorem Matters for JEE
The Multinomial Theorem extends the familiar Binomial Theorem to multiple variables, providing powerful tools for:
- Complex expansions of $(x_1 + x_2 + \cdots + x_k)^n$
- Coefficient problems in JEE Advanced
- Combinatorial identities and proofs
- Probability applications with multiple outcomes
The Multinomial Theorem Formula
For any positive integer $n$ and any non-negative integers $n_1, n_2, \ldots, n_k$ such that $n_1 + n_2 + \cdots + n_k = n$:
$$(x_1 + x_2 + \cdots + x_k)^n = \sum \frac{n!}{n_1! n_2! \cdots n_k!} x_1^{n_1} x_2^{n_2} \cdots x_k^{n_k}$$
where the sum is over all k-tuples of non-negative integers $(n_1, n_2, \ldots, n_k)$ with $\sum n_i = n$
Understanding the Formula:
Multinomial Coefficient: $\frac{n!}{n_1! n_2! \cdots n_k!}$ counts the number of ways to arrange $n$ objects where there are $n_1$ of type 1, $n_2$ of type 2, etc.
Connection to Binomial: When $k=2$, this reduces to the familiar binomial theorem with $\frac{n!}{n_1! n_2!} = \binom{n}{n_1}$
Basic Expansion Examples
Example 1: Expand $(x + y + z)^2$
All possible combinations where $n_1 + n_2 + n_3 = 2$:
Case 1: $(2,0,0)$: $\frac{2!}{2!0!0!}x^2 = x^2$
Case 2: $(0,2,0)$: $\frac{2!}{0!2!0!}y^2 = y^2$
Case 3: $(0,0,2)$: $\frac{2!}{0!0!2!}z^2 = z^2$
Case 4: $(1,1,0)$: $\frac{2!}{1!1!0!}xy = 2xy$
Case 5: $(1,0,1)$: $\frac{2!}{1!0!1!}xz = 2xz$
Case 6: $(0,1,1)$: $\frac{2!}{0!1!1!}yz = 2yz$
Result: $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz$
Example 2: Expand $(a + b + c)^3$
All combinations where $n_1 + n_2 + n_3 = 3$:
• $(3,0,0)$: $a^3$
• $(0,3,0)$: $b^3$
• $(0,0,3)$: $c^3$
• $(2,1,0)$: $3a^2b$
• $(2,0,1)$: $3a^2c$
• $(1,2,0)$: $3ab^2$
• $(0,2,1)$: $3b^2c$
• $(1,0,2)$: $3ac^2$
• $(0,1,2)$: $3bc^2$
• $(1,1,1)$: $6abc$
Result: $(a + b + c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc$
Coefficient Problems in JEE
Example 3: Find coefficient of $x^2y^3z^2$ in $(x + y + z)^7$
Step 1: Identify exponents: $n_1=2$, $n_2=3$, $n_3=2$
Step 2: Check sum: $2 + 3 + 2 = 7$ ✓
Step 3: Apply multinomial coefficient:
$\frac{7!}{2!3!2!} = \frac{5040}{2 \cdot 6 \cdot 2} = \frac{5040}{24} = 210$
Answer: Coefficient is 210
Example 4: Find coefficient of $a^3b^2c^4$ in $(2a - b + 3c)^9$
Step 1: Identify exponents: $n_1=3$, $n_2=2$, $n_3=4$
Step 2: Check sum: $3 + 2 + 4 = 9$ ✓
Step 3: Apply multinomial coefficient:
$\frac{9!}{3!2!4!} = \frac{362880}{6 \cdot 2 \cdot 24} = \frac{362880}{288} = 1260$
Step 4: Include coefficient factors: $(2)^3 \cdot (-1)^2 \cdot (3)^4 = 8 \cdot 1 \cdot 81 = 648$
Step 5: Multiply: $1260 \times 648 = 816,480$
Answer: Coefficient is 816,480
🚀 Problem-Solving Strategies
For Coefficient Problems:
- Always verify $n_1 + n_2 + \cdots + n_k = n$
- Remember to include coefficient factors from each term
- Watch for negative signs in the expansion
- Use symmetry when possible to reduce calculations
For Special Cases:
- When finding sum of coefficients, set all variables = 1
- For specific term problems, use the general term formula
- Remember the multinomial coefficient counts arrangements
- Practice with substitution methods for complex problems
Advanced Applications Available in Full Version
Includes Number of Terms analysis, Relationship with Permutations, and JEE Advanced level problems
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Find the coefficient of $x^3y^2z$ in $(x + y + z)^6$
2. How many terms are in the expansion of $(x + y + z + w)^5$?
3. Find the sum of all coefficients in $(2x - 3y + z)^4$
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