Key Concept: The Multinomial Theorem

This problem asks for the coefficient of a specific term ($x^4$) in the expansion of a trinomial $(1 + x + x^2)$ raised to a power ($10$). The most direct and robust method for solving such problems is by utilizing the Multinomial Theorem.

The Multinomial Theorem provides a general formula for expanding expressions of the form $(a_1 + a_2 + \ldots + a_m)^n$. The general term in this expansion is given by: $\frac{n!}{k_1! k_2! \ldots k_m!} a_1^{k_1} a_2^{k_2} \ldots a_m^{k_m}$ where $k_1, k_2, \ldots, k_m$ are non-negative integers such that their sum equals $n$ (i.e., $k_1 + k_2 + \ldots + k_m = n$).

In our specific problem, we have $(1 + x + x^2)^{10}$. Here, the terms are $a_1=1$, $a_2=x$, and $a_3=x^2$, and the power $n=10$. Let the powers of $1$, $x$, and $x^2$ in a general term be $p$, $q$, and $r$ respectively. So, $k_1=p$, $k_2=q$, and $k_3=r$.

The general term in the expansion of $(1 + x + x^2)^{10}$ is: $\frac{10!}{p!q!r!} (1)^p (x)^q (x^2)^r$ Simplifying the powers of $x$: $\frac{10!}{p!q!r!} x^{q+2r}$ The conditions for $p, q, r$ are:

1. $p, q, r$ must be non-negative integers ($p \ge 0, q \ge 0, r \ge 0$).
2. The sum of their powers must equal the overall exponent: $p + q + r = 10$.

Step-by-step Working: Finding the Coefficient of $x^4$

We need to find the coefficient of $x^4$. This means the exponent of $x$ in the general term must be equal to 4. So, our primary condition is: $q + 2r = 4$ Now, we need to find all possible combinations of non-negative integers $(p, q, r)$ that satisfy both $q+2r=4$ and $p+q+r=10$. We will systematically explore values for $r$, starting from $0$.

Case 1: When $r = 0$

• From $q + 2r = 4$: $q + 2(0) = 4 \Rightarrow q = 4$.
• From $p + q + r = 10$: $p + 4 + 0 = 10 \Rightarrow p = 6$.
• This gives us the triplet $(p,q,r) = (6,4,0)$.
• The coefficient for this term is $\frac{10!}{6!4!0!}$. (Remember that $0! = 1$) $\frac{10 \times 9 \times 8 \times 7 \times 6!}{6! \times (4 \times 3 \times 2 \times 1) \times 1} = \frac{10 \times 9 \times 8 \times 7}{24} = 10 \times 3 \times 7 = 210$

Case 2: When $r = 1$

• From $q + 2r = 4$: $q + 2(1) = 4 \Rightarrow q + 2 = 4 \Rightarrow q = 2$.
• From $p + q + r = 10$: $p + 2 + 1 = 10 \Rightarrow p + 3 = 10 \Rightarrow p = 7$.
• This gives us the triplet $(p,q,r) = (7,2,1)$.
• The coefficient for this term is $\frac{10!}{7!2!1!}$. $\frac{10 \times 9 \times 8 \times 7!}{7! \times (2 \times 1) \times 1} = \frac{10 \times 9 \times 8}{2} = 10 \times 9 \times 4 = 360$

Case 3: When $r = 2$

• From $q + 2r = 4$: $q + 2(2) = 4 \Rightarrow q + 4 = 4 \Rightarrow q = 0$.
• From $p + q + r = 10$: $p + 0 + 2 = 10 \Rightarrow p + 2 = 10 \Rightarrow p = 8$.
• This gives us the triplet $(p,q,r) = (8,0,2)$.
• The coefficient for this term is $\frac{10!}{8!0!2!}$. $\frac{10 \times 9 \times 8!}{8! \times 1 \times (2 \times 1)} = \frac{10 \times 9}{2} = 45$

Case 4: When $r \ge 3$ If we try $r=3$, then $q + 2(3) = 4 \Rightarrow q + 6 = 4 \Rightarrow q = -2$. Since $q$ must be a non-negative integer, we cannot have $r \ge 3$. Thus, we have considered all possible valid combinations.

Total Coefficient Calculation The total coefficient of $x^4$ is the sum of the coefficients from all valid cases: $\text{Coefficient of } x^4 = (\text{Case 1}) + (\text{Case 2}) + (\text{Case 3})$ $\text{Coefficient of } x^4 = 210 + 360 + 45 = 615$

Alternative Approach: Repeated Binomial Expansion

One can also solve this problem by applying the Binomial Theorem iteratively. We can treat $(1+x+x^2)^{10}$ as $[(1+x) + x^2]^{10}$. Using the binomial theorem $(A+B)^n = \sum_{k=0}^{n} \binom{n}{k} A^{n-k} B^k$, with $A=(1+x)$ and $B=x^2$: $( (1+x) + x^2 )^{10} = \binom{10}{0} (1+x)^{10} (x^2)^0 + \binom{10}{1} (1+x)^9 (x^2)^1 + \binom{10}{2} (1+x)^8 (x^2)^2 + \binom{10}{3} (1+x)^7 (x^2)^3 + \ldots$ We need to find terms that contribute to $x^4$:

1. From $\binom{10}{0} (1+x)^{10}$: We need the coefficient of $x^4$ in $(1+x)^{10}$, which is $\binom{10}{4}$. $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

2. From $\binom{10}{1} (1+x)^9 (x^2)^1 = 10 x^2 (1+x)^9$: We already have an $x^2$ term. To get $x^4$ overall, we need an $x^2$ term from $(1+x)^9$. The coefficient of $x^2$ in $(1+x)^9$ is $\binom{9}{2}$. $\binom{10}{1} \times \binom{9}{2} = 10 \times \frac{9 \times 8}{2 \times 1} = 10 \times 36 = 360$

3. From $\binom{10}{2} (1+x)^8 (x^2)^2 = 45 x^4 (1+x)^8$: We already have an $x^4$ term. To get $x^4$ overall, we need the constant term (coefficient of $x^0$) from $(1+x)^8$. The coefficient of $x^0$ in $(1+x)^8$ is $\binom{8}{0}$. $\binom{10}{2} \times \binom{8}{0} = 45 \times 1 = 45$

4. From $\binom{10}{3} (1+x)^7 (x^2)^3 = \binom{10}{3} (1+x)^7 x^6$: This term already contains $x^6$, which is a higher power than $x^4$. Therefore, this term and any subsequent terms will not contribute to the coefficient of $x^4$.

Summing the contributions: $210 + 360 + 45 = 615$. Both methods confirm the result.

Tips and Common Mistakes

• Systematic Case Enumeration: When using the Multinomial Theorem, always systematically list all possible non-negative integer combinations of powers $(p, q, r)$ that satisfy both the sum constraint ($p+q+r=n$) and the target power constraint ($q+2r=4$). This prevents missing terms.
• Factorial Calculations: Be meticulous when calculating factorial expressions for multinomial coefficients. Remember $0! = 1$.
• Power of $x$ in terms: Pay close attention to how the powers of $x$ combine. In $(x^2)^r$, the power of $x$ is $2r$, not just $r$.
• Binomial vs. Multinomial: For trinomials or larger polynomials, the Multinomial Theorem is generally more elegant and less prone to errors than repeated binomial expansion.

Summary and Key Takeaway

To find the coefficient of $x^4$ in the expansion of $(1 + x + x^2)^{10}$, we applied the Multinomial Theorem. By identifying all combinations of non-negative integer powers $(p, q, r)$ for $1, x, x^2$ respectively, such that their sum is $10$ and the total power of $x$ is $4$, we calculated the individual coefficients for each combination. Summing these individual coefficients ($210 + 360 + 45$) yielded the final answer of $615$. This problem is a good illustration of how combinatorial principles are applied in polynomial expansions.