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\begin{align}
&\pars{1 + x + x^{2}}\pars{1 − x}^{8} =
\pars{1 - x^{3}}\pars{1 - x}^{7} =
\sum_{n = 0}^{7}{7 \choose n}\pars{-1}^{n}x^{n} -
\sum_{n = 0}^{7}{7 \choose n}\pars{-1}^{n}x^{n + 3}
\\[5mm] = &\
\sum_{n = 0}^{7}{7 \choose n}\pars{-1}^{n}x^{n} +
\sum_{n = 3}^{10}{7 \choose n - 3}\pars{-1}^{n}x^{n}
\\[5mm] = &\
\bracks{1 - 7x + 21x^{2} + \sum_{n = 3}^{7}{7 \choose n}\pars{-1}^{n}x^{n}} +
\bracks{\sum_{n = 3}^{7}{7 \choose n - 3}\pars{-1}^{n}x^{n} + 21x^{8} - 7x^{9} + x^{10}}
\\[5mm] = &\
1 - 7x + 21x^{2} +
\sum_{n = 3}^{7}\bracks{{7 \choose n} + {7 \choose n - 3}}\pars{-1}^{n}x^{n} +
21x^{8} - 7x^{9} + x^{10} = \sum_{n = 0}^{10}a_{n}x^{n}
\end{align}
where
$$
a_{n} =
\left\{\begin{array}{lcl}
\ds{\phantom{-}1} & \mbox{if} & \ds{n = 0}
\\[2mm]
\ds{-7} & \mbox{if} & \ds{n = 1}
\\[2mm]
\ds{\phantom{-}21} & \mbox{if} & \ds{n = 2}
\\[2mm]
\ds{\pars{-1}^{n}\bracks{%
{7 \choose n} + {7 \choose n - 3}}} & \mbox{if} & \ds{3 \leq n \leq 7}
\\[2mm]
\ds{a_{10 -n}} & \mbox{if} & \ds{8 \leq n \leq 10}
\end{array}\right.
$$
Indeed, the coefficients $\ds{\braces{a_{n}}}_{\ 0\ \leq\ n\ \leq\ 10}$ satisfacen $\ds{a_{n} = a_{10 - n}}$ such that's sufficient to enumerate the values of $\ds{a_{n}}$ for $\ds{n = 0,1,\ldots,5}$. Namely,
$$
a_{0} = 1\,,\ a_{1} = -7\,,\ a_{2} = 21\,, a_{3} = -36\,,\ a_{4} = 42\,,\
a_{5} = -42
$$
To convert to a more familiar form, multiply and divide by $x^6$:
$$(x-1)^3 (1+2x^2)^6 x^{-6}$$
Now you're looking for the coefficient of $x^4$ in $(x-1)^3 (1+2x^2)^6$. You can get $x^4$ out of this product either by multiplying $x^2$ with $x^2$ or $1$ with $x^4$. Each of those coefficients should be easy to calculate, then you just add them.
Best Answer
There are 101 terms in the first factor, of which 100 have a power of $x$ greater than 0. The second factor can be written:
$$ \sum_{k_1 + k_2 + k_3 = 101} 1^{k_1} \cdot x^{2 k_2} \cdot x^{k_3} $$
(just imagine the product multiplied out) so you are asking how many sets of (k_1, k_2, k_3), all integers at least 0, are such that $k_1 + k_2 + k_3 = 101$ with $k_1 < 101$ (that one gives the term 1, there is just one of those).
Let's find out how many solutions $k_1 + k_2 + k_3 = 101$ has, this is like chopping a line of 101 $*$ into three pieces, say by separating with $|$ (this is called a stars and bars argument, for obvious reasons). But then you have a total of $101 + 2$ positions to be filled with 101 stars and 2 bars, that can be done in $\binom{101 + 2}{2} = 5253$ ways, of which you subtract 1 for $k_1 = 101$.
Combining your factors, you have 101 terms in the first factor, $5252$ with $x$ from the second, and the terms with $x$ in the product are the result of multiplying any term from the first factor with a term containing $x$ from the second, i.e., $101 \cdot 5252 = 53042$ (as long as no simpolifications happen).
A lower bound is that the result is a polynomial of degree $100 + 2 \cdot 101 = 302$, so there are at most $301$ terms with powers of $x$. But there are negative terms, so cancellation can/will happen, and you get less.