Find the coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$ The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$.
I tried the following concept, how to sum 9 using 1-9 only without repetition.
1:9, 2:1+8,3:7+2, 4: 6+3, 5: 5+4,
6:1+2+6 ,7:1+3+5, 8:2+3+4
The answer is 8.
How will it be solved using Binomial theorem.
 A: Yes we can count directly the cases that is


*

*$x^9$

*$x\cdot x^8$

*$x\cdot x^2\cdot x^6$

*$x\cdot x^3\cdot x^5$

*$x^2\cdot x^7$

*$x^2\cdot x^3\cdot x^4$

*$x^3\cdot x^6$

*$x^4\cdot x^5$


to obtain $8$.
A: As Theo Bendit has already commented this is not solved by Newton's binomial theorem. The solution is the same way that is used to prove Newton's bionomian theorem that is the proper use of combinatorial principles. Let
$$
p(x)=(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)(1+x^6)(1+x^7)(1+x^8)(1+x^9)
$$
We have
\begin{align}
p(x)
=& 
1
+\sum_{1\leq i_1\leq 9}x^{i_1}
+\sum_{1\leq i_1<i_2\leq 9}x^{i_1}\cdot x^{i_2}
+\sum_{1\leq i_1<i_2<i_3\leq 9}x^{i_1}\cdot x^{i_2}\cdot x^{i_3}
+
\\
\\
&\qquad
+\sum_{1\leq i_1<i_2<i_3<i_4\leq 9}x^{i_1}\cdot x^{i_2}\cdot x^{i_3}\cdot x^{i_4} 
+\ldots +\sum_{1\leq i_1<\cdots <i_9\leq 9}x^{i_1}\cdot \dots \cdot x^{i_9}
\end{align}
or for $m_1=\max\{i_1: 1\leq i_1\leq 9\}$, $m_2=\max\{i_1+i_2: : 1\leq i_1<i_2\leq 9\}$,$\ldots$, $m_9=\max\{i_1+\ldots+i_9: 1\leq i_1<\ldots <i_9\leq 9\}$
\begin{align}
p(x)
=& 
1
+\sum_{k=1}^{m_1}\sum_{i_1= k}x^{i_1}
+\sum_{k=1+2}^{m_2}\sum_{\substack{ i_1<i_2 \\i_1+i_2=k}}x^{i_1}\cdot x^{i_2}
+\sum_{k=1+2+3}^{m_3}\sum_{\substack{ i_1<i_2<i_3 \\i_1+i_2+i_3=k}}x^{i_1}\cdot x^{i_2}\cdot x^{i_3}
+\ldots
\\
\\
&\qquad
+\sum_{k=1+2+3+4}^{m_4}\;\;\sum_{i_1<i_2<i_3<i_4}x^{i_1}\cdot x^{i_2}\cdot x^{i_3}\cdot x^{i_4} 
+
\qquad\ldots+\sum_{k=1+\ldots+9}^{m_9}\sum_{\substack{ i_1<\cdots <i_9\\ i_1+\ldots+i_9=k}}x^{i_1}\cdot \dots \cdot x^{i_9}
\end{align}
All possible powers of $ x $ equal to $ x ^ 9 $ are generated by the first three summations.
Let $ c_9 $ be the coefficient of $ x ^ 9 $ in the polynomial $ p (x) $. So
\begin{align}
c_9\cdot x^9 
=& 
\sum_{\substack{1\leq i_1\leq 9\\ i_1=9}}x^{i_1}
+\sum_{\substack{1\leq i_1<i_2\leq 9\\ i_1+i_2=9}}x^{i_1}\cdot x^{i_2}
+\sum_{\substack{1\leq i_1<i_2<i_3\leq 9\\ i_1+i_2+i_3=9}}x^{i_1}\cdot x^{i_2}\cdot x^{i_3}
\\
=& 1\cdot x^9+1\cdot x^1\cdot x^8+1\cdot x^2\cdot x^7+1\cdot x^3\cdot x^6+1\cdot x^4\cdot x^5
\\
&+1\cdot x^1\cdot x^2\cdot x^6+1\cdot x^1\cdot x^3\cdot x^5+1\cdot x^2\cdot x^3\cdot x^4
\\
&=8x^9
\end{align}
