Number of polynomials of degree 4 Find the numbers of all polynomials of degree $4$ you can create with the letters $x,y,z$. All coefficients must be $1$.
I can create $\binom{6}{4}=15$ monomials of degree $4$ with $x,y,z$. Using the same rule I obtain $10,6,3,1$ monomials of degree $3,2,1,0$ respectively with the same letters. Now Call $A$ the set made of all subsets of monomials of degree $4$ (except for the null set), $|A|=2^{15}-1$, and call $B$ the set made by all subsets of monomials of degree from $3$ to $0$, $|B|=2^{20}$. All polynomials of degree $4$ are $$|A \times B|=(2^{15}-1)2^{20}=34358689792$$ which is huge. Is my answer correct?
 A: Your answer looks correct to me.
A: Let me present a slight different way to count such polynomials.
Consider the sets of the monomials
$$
S_{\,4}  = \underbrace {\left\{ {x^{\,n_{\,x} } y^{\,n_{\,y} } z^{\,n_{\,z} } } \right\}}_{\left\{ \begin{subarray}{l} 
  0\, \leqslant \,n_{\,k} \, \leqslant \,4\, \\ 
  n_{\,x}  + n_{\,y}  + n_{\,z}  = 4 
\end{subarray}  \right.},\quad S_{\,3}  = \underbrace {\left\{ {x^{\,n_{\,x} } y^{\,n_{\,y} } z^{\,n_{\,z} } } \right\}}_{\left\{ \begin{subarray}{l} 
  0\, \leqslant \,n_{\,k} \, \leqslant \,3\, \\ 
  n_{\,x}  + n_{\,y}  + n_{\,z}  = 3 
\end{subarray}  \right.},\; \cdots \;,\quad S_{\,0}  = \underbrace {\left\{ {x^{\,n_{\,x} } y^{\,n_{\,y} } z^{\,n_{\,z} } } \right\}}_{\left\{ \begin{subarray}{l} 
  0\, \leqslant \,n_{\,k} \, \leqslant \,0\, \\ 
  n_{\,x}  + n_{\,y}  + n_{\,z}  = 0 
\end{subarray}  \right.}
$$
The cardinality of the set $S_m$ will then correspond to the number of weak compositions of $m$ into $3$ parts, i.e.: 
$$
\begin{gathered}
  S_{\,m}  = \underbrace {\left\{ {x^{\,n_{\,x} } y^{\,n_{\,y} } z^{\,n_{\,z} } } \right\}}_{\left\{ \begin{subarray}{l} 
  0\, \leqslant \,n_{\,k} \, \leqslant \,m\, \\ 
  n_{\,x}  + n_{\,y}  + n_{\,z}  = m 
\end{subarray}  \right.}\quad  \Rightarrow \quad \left| {S_{\,m} } \right| = \left( \begin{gathered}
  m + 3 - 1 \\ 
  3 - 1 \\ 
\end{gathered}  \right) = \left( \begin{gathered}
  m + 2 \\ 
  2 \\ 
\end{gathered}  \right) =  \hfill \\
   = \left[ {1,3,6,10,15\;\left| {\;m = 0, \cdots ,4} \right.} \right] \hfill \\ 
\end{gathered} 
$$
and the cardinality of the total set, union of the sets above, will be
$$
S = \bigcup\limits_{0\, \le \,m\, \le \,4} {S_{\,m} } \quad  \Rightarrow \quad \left| S \right| = \sum\limits_{0\, \le \,m\, \le \,4} {\left| {S_{\,m} } \right|}  = \sum\limits_{0\, \le \,m\, \le \,4} {\left( \matrix{
  m + 2 \cr 
  2 \cr}  \right)}  = \left( \matrix{
  7 \cr 
  4 \cr}  \right) = 35
$$
Now a polynomial of degree $4$ shall be made by the sum of at least one of the monomials in $S_4$ and of whichever
number of the monomials of lower degree, thus
$$
N = \left( {2^{\,15}  - 1} \right)2^{\,20} 
$$
which confirms your calculation.
