Coefficients..........

(a + a^2 + a^3)^4=a^12+4 a^11+10 a^10+16 a^9+19 a^8+16 a^7+10 a^6+4 a^5+a^4
In the expanded part, the term "19a^8", how is the coefficient 19 arrived at? Thanks for any help.

Formula trinomial:

$\begin{array}{|rcll|} \hline (a+b+c)^n &=& \sum \limits_{i=0}^{n} \sum \limits_{j=0}^{n-i} \frac{n!}{i!~j!~(n-i-j)!} \cdot a^ib^jc^{n-i-j} \\ \hline \end{array}$

We have:

$\begin{array}{|rcll|} \hline (a + a^2 + a^3)^4 &=& \sum \limits_{i=0}^{4} \sum \limits_{j=0}^{4-i} \frac{4!}{i!~j!~(4-i-j)!} \cdot a^i(a^2)^j(a^3)^{4-i-j} \\ &=& \sum \limits_{i=0}^{4} \sum \limits_{j=0}^{4-i} \frac{4!}{i!~j!~(4-i-j)!} \cdot a^{i+2j+3\cdot(4-i-j)} \\ &=& \sum \limits_{i=0}^{4} \sum \limits_{j=0}^{4-i} \frac{4!}{i!~j!~(4-i-j)!} \cdot a^{12-2i-j} \\ \hline \end{array}$

We need the coefficient $c_8$ of $a^8$:

So

$\begin{array}{|rcll|} \hline a^{12-2i-j} &=& a^8 \\ 12-2i-j &=& 8 \\ \mathbf{j} &\mathbf{=}& \mathbf{4-2i}\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline c_8\cdot a^8 &=& \sum \limits_{i=0}^{4} \sum \limits_{j=0}^{4-i} \frac{4!}{i!~j!~(4-i-j)!} \cdot a^{8} \quad | \quad j=4-2i\\ c_8\cdot a^8 &=& a^{8} \cdot \sum \limits_{i=0}^{4} \frac{4!}{i!~(4-2i)!~[4-i-(4-2i)]!} \\ c_8\cdot a^8 &=& a^{8} \cdot \sum \limits_{i=0}^{4} \frac{4!}{i!~(4-2i)!~i!} \\ && 4-2i \ge 0 \\ && 4 \ge 2i \\ && 2 \ge i \\ && \mathbf{i \le 2}\\ c_8\cdot a^8 &=& a^{8} \cdot \sum \limits_{i=0}^{2} \frac{4!}{i!~(4-2i)!~i!} \\ c_8\cdot a^8 &=& a^{8} \cdot \left[ \frac{4!}{0!~(4-2\cdot 0)!~0!} +\frac{4!}{1!~(4-2\cdot 1)!~1!} +\frac{4!}{2!~(4-2\cdot(2))!~2!} \right] \\ c_8\cdot a^8 &=& a^{8} \cdot ( \frac{4!}{4!} +\frac{4!}{2!} +\frac{4!}{2!~2!} ) \\ c_8\cdot a^8 &=& a^{8} \cdot ( 1 +3\cdot 4 +\frac{3\cdot 4}{2} ) \\ c_8\cdot a^8 &=& a^{8} \cdot ( 1 +12 + 6 ) \\ c_8\cdot a^8 &=& a^{8} \cdot 19 \\ c_8\cdot a^8 &=& \mathbf{19}\cdot a^{8} \\ \hline \end{array}$