Find the coefficient of x⁴ in (1 + x + x²)¹⁰

Q: Why can't I just use the Binomial Theorem formula $(1+x)^n$ by grouping terms?

That is a common temptation! You could group $(1+x+x^2)$ as $(1+(x+x^2))$, but that would require expanding a binomial to the power of $10$, and then expanding the resulting $(x+x^2)^k$ terms. It creates a nested, messy calculation that is prone to error. The Multinomial Theorem is the direct, elegant path designed specifically for this structure. It treats all three terms symmetrically, which is mathematically more robust.

Q: How do I know when to stop checking cases for $\gamma$?

Look at the constraint $\beta + 2\gamma = 4$. Since $\beta$ must be a non-negative integer, $2\gamma$ cannot exceed $4$. Therefore, $\gamma$ can only be $0, 1,$ or $2$. Once $\gamma$ hits $3$, $2\gamma$ becomes $6$, which forces $\beta$ to be $-2$. Since we cannot choose a term a negative number of times, we stop immediately. The constraints define the boundaries of your search space.

Q: Is $0!$ really $1$? I always feel like it should be $0$.

It is a classic point of confusion! In combinatorics, $0!$ is defined as $1$ because there is exactly one way to arrange zero items: by doing nothing. If you treat it as $0$, your entire calculation will collapse into zero. Always remember this fundamental definition; it is the bedrock upon which these coefficients are built.

Question

Find the coefficient of $x^4$ in $(1 + x + x^2)^{10}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The expression $(1 + x + x^2)^{10}$ presents a classic challenge in combinatorics. While it resembles a binomial expansion, the presence of three terms necessitates the use of the **Multinomial Theorem**.

This theorem provides a systematic way to determine the coefficient of any specific power of $x$ within the expansion. We are essentially counting the number of ways to select terms from ten brackets such that their product results in $x^4$.

### The General Term

The Multinomial Theorem states that the general term in the expansion of $(a + b + c)^n$ is given by:

$$T = \frac{n!}{\alpha! \beta! \gamma!} a^\alpha b^\beta c^\gamma$$

In our specific case, $n = 10$, $a = 1$, $b = x$, and $c = x^2$. Substituting these values, the general term becomes:

$$T = \frac{10!}{\alpha! \beta! \gamma!} (1)^\alpha (x)^\beta (x^2)^\gamma = \frac{10!}{\alpha! \beta! \gamma!} x^{\beta + 2\gamma}$$

Our objective is to find the coefficient of $x^4$. Therefore, we must satisfy the condition $\beta + 2\gamma = 4$.

### The Constraints

To solve for the variables $\alpha$, $\beta$, and $\gamma$, we must satisfy two fundamental constraints derived from the expansion:

1. The sum of the exponents must equal the total power: $\alpha + \beta + \gamma = 10$.
2. The total power of $x$ must be four: $\beta + 2\gamma = 4$.

We seek all non-negative integer triplets $(\alpha, \beta, \gamma)$ that satisfy these equations. We will iterate through possible values of $\gamma$ to find valid solutions.

### The Detective Work

**Case 1: Let $\gamma = 0$**

Substituting $\gamma = 0$ into the power constraint $\beta + 2(0) = 4$ yields $\beta = 4$. Using the sum constraint, $\alpha + 4 + 0 = 10$, we find $\alpha = 6$.

The triplet is $(6, 4, 0)$. The coefficient contribution is:

$$\frac{10!}{6! 4! 0!} = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$$

**Case 2: Let $\gamma = 1$**

Substituting $\gamma = 1$ into the power constraint $\beta + 2(1) = 4$ yields $\beta = 2$. Using the sum constraint, $\alpha + 2 + 1 = 10$, we find $\alpha = 7$.

The triplet is $(7, 2, 1)$. The coefficient contribution is:

$$\frac{10!}{7! 2! 1!} = \frac{10 	imes 9 	imes 8}{2 	imes 1} = 360$$

**Case 3: Let $\gamma = 2$**

Substituting $\gamma = 2$ into the power constraint $\beta + 2(2) = 4$ yields $\beta = 0$. Using the sum constraint, $\alpha + 0 + 2 = 10$, we find $\alpha = 8$.

The triplet is $(8, 0, 2)$. The coefficient contribution is:

$$\frac{10!}{8! 0! 2!} = \frac{10 	imes 9}{2 	imes 1} = 45$$

### Final Calculation

We have exhausted all possible non-negative integer solutions, as any $\gamma \ge 3$ would result in a negative value for $\beta$. To find the total coefficient of $x^4$, we sum the contributions from each valid case:

$$	ext{Total Coefficient} = 210 + 360 + 45 = 615$$

The final coefficient of $x^4$ in the expansion of $(1 + x + x^2)^{10}$ is **615**.

Animated Solution for Mathematics - Binomial Theorem: Find the coefficient of $x^{4}$ in $(1 + x + x^{2})^{10}$

Enter Numerical Value:

Visualized Solution

Identifying the Multinomial Expression

Setting up the General Term

Simplifying the Power of $x$

Case 1: When $γ = 0$

Calculating Coefficient for Case 1

Case 2: When $γ = 1$

Calculating Coefficient for Case 2

Case 3: When $γ = 2$

Calculating Coefficient for Case 3

Summing Up the Results

Summary and Key Takeaway

The Sigma Insight: Multinomial Theorem

Analyzing the Setup

The General Term

The Constraints

The Detective Work

Final Calculation

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the coefficient of x4 in (1+x+x2)10

Enter Numerical Value:

Visualized Solution

Identifying the Multinomial Expression

Setting up the General Term

Simplifying the Power of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Case 1: When .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ=0