<s>[INST] Number of positive unequal integer solutions of $x+y+z+w=20$ What is the number of positive different integer solutions of $x+y+z+w=20$, where $x,y,z,w$ are all different and positive?
It would be nice if coding is not used.  I am given the answer $552$.
 [/INST] By inclusion-exclusion over the distinctness conditions:
There are $\binom{20-1}3=\binom{19}3$ solutions disregarding the conditions.
There are $\binom42$ ways to choose one pair that fails to be distinct, with common value $k$, and $19-2k$ options for the remaining two values, for a total of
\begin{align}
\sum_{k=1}^9(19-2k)&=9\cdot19-2\cdot\frac{9(9+1)}2\\
&=81\;.
\end{align}
There are $3$ ways to choose two non-overlapping pairs that fail to be distinct, and $\binom{10-1}1=9$ options for their values, and $12$ ways to choose two overlapping pairs that fail to be distinct, so that three values are equal, and then $6$ different options for the values.
There are $\binom{\binom42}3=\binom63$ ways to choose three pairs, of which $4$ constrain only $3$ values to be equal, leading to $6$ options as above, and the remaining $\binom63-4$ constrain all $4$ values to be equal, leaving only $1$ option.
For $k=4,5,6$, there are $\binom{\binom42}k=\binom6k$ ways to choose $k$ pairs, all of which constrain all $4$ values to be equal and thus leave only $1$ option.
Thus by inclusion-exclusion the total count is
$$
\binom{19}3-\binom42\cdot81+3\cdot9+12\cdot6-4\cdot6-\left(\binom63-4\right)+\binom64-\binom65+\binom66=552\;.
$$
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