# Multinomial Verification

Here's my question, which is similar to the one I asked a couple weeks ago:

Find the coefficient of $x^4$ in the expansion of $(1+x+x^2+x^3)^7$.

I was suggested to use the Binomial Theorem, but it would become very tedious and just decided to use the Multinomial Theorem instead. So I need to find the terms that have $x^4$ in them. 3 such instances do occur and they are:

$$1^5 \cdot (x)^1 \cdot (x^2)^0 \cdot (x^3)^1$$ $$1^5 \cdot (x)^0 \cdot (x^2)^2 \cdot (x^3)^0$$ $$1^3 \cdot (x)^4 \cdot (x^2)^0 \cdot (x^3)^0$$

Notice that all of the exponents add up to $7$. I'm supposed to get $203$ as my result (verified using Wolfram), but I got $98$ as my answer. I believe I'm missing some other factors, but cannot find any more other than those $3$.

$$\begin{eqnarray*}[x^4](1+x+x^2+x^3)^7 &=& [x^4](1-x^4)^7\cdot\frac{1}{(1-x)^7}\\&=&[x^4]\left(\sum_{k=0}^{7}(-1)^k \binom{7}{k}x^{4k}\right)\cdot\left(1 + 7 x + 28 x^2 + 84 x^3 + 210 x^4+\ldots\right)\end{eqnarray*}$$ hence the answer is given by $\binom{7}{0}\cdot 210-\binom{7}{1}\cdot 1 = \color{red}{203}$ as predicted by WA.
• @i8Σπ_821: I am considering the Cauchy product between two well-known Taylor series and extracting the coefficient of $x^4$, that we may only get as $x^0\cdot x^4$ or as $x^4\cdot x^0$. – Jack D'Aurizio Mar 13 '17 at 20:04
You are missing the combination: $$1^4 \cdot (x)^2 \cdot (x^2)^1 \cdot (x^3)^0 \to \frac{7!}{4! \, 2! \,1!\, 0!}=105$$