We are given the task of placing 7 balls into 3 jars. Step 1: Place 1st ball, 3 ways to do that. Step 2: Place 2nd ball, 3 ways to do that....Step 7: place last(seventh) ball, 3 ways to do that. By rule of product, we have $3*3*3*3*3*3*3 = 3^7$ ways to accomplish the task. Your method is wrong because assumes we need to put a ball in the first jar. We don't need to put anything in the first jar.
I'll solve the case in which the balls are identical. This is not the nicest solution, but it works and I hope it is clear.
Case 1: First two boxes have exactly four balls.
If the first two boxes together contain four balls, then we can use stars and bars twice. The number of ways to place four balls in two boxes is the number of ways to arrange $4$ stars and $1$ bar, or $5$ choose $1$. The number of ways to place the remaining four balls in the other four boxes is the number of was to arrange $4$ stars and $3$ bars, or $7$ choose $3$. This gives
$$\binom{5}{1}\binom{7}{3}$$
Case 2: First two boxes have exactly three balls. With the same reasoning, we get
$$\binom{4}{1}\binom{8}{3}$$
Case 3: First two boxes have exactly two balls.
$$\binom{3}{1} \binom{9}{3}$$
Case 4: First two boxes have exactly one ball.
$$\binom{2}{1}\binom{10}{3}$$
Case 5: First two boxes have no balls.
$$\binom{1}{1}\binom{11}{3}$$
(This makes sense because this is just stars and bars applied to putting $8$ balls into $4$ boxes)
Our total is therefore
$$5\binom{7}{3}+4\binom{8}{3}+3\binom{9}{3}+2\binom{10}{3}+\binom{11}{3}=\boxed{1056}$$
Best Answer
Let $g_{e}(x)=\displaystyle\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right)^3\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)^2=(e^x)^3\left(\frac{e^x+e^{-x}}{2}\right)^2$
$\;\;\;\;\;\;\;\;\;\;\;\;=\displaystyle e^{3x}\left(\frac{e^{2x}+2+e^{-2x}}{4}\right)=\frac{1}{4}(e^{5x}+2e^{3x}+e^x)$.
The coefficient of $x^{100}$ is given by $\displaystyle\frac{a_{100}}{100!}=\frac{1}{4}\left(\frac{5^{100}}{100!}+2\cdot\frac{3^{100}}{100!}+\frac{1}{100!}\right)$, so
$\hspace{1.5 in}$ $\displaystyle a_{100}=\frac{1}{4}(5^{100}+2\cdot3^{100}+1)$.