How much ways there are to produce sum = 21 with 4 different natural numbers? 0 isn't natural number , and the sum way is important (e.g. 3+5+6+7 is different from 5+6+7+3).
I got that I have for a+b+c+d = 21 , 2024 options.
I think I need to sub the invalid numbers , so I need to relate for the cases :
a = 0 , b = 0, c = 0, d = 0 , 
a = b , a = c , a = d, b = c, b = d, c = d?

$$
\eqalign{
  & N = {\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {a + b + c + d = 21\quad \left| {\;0 < a \ne b \ne c \ne d} \right.} \right)  \cr 
  & \quad  \Downarrow   \cr 
  & N = 4!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {a + b + c + d = 21\quad \left| {\;0 < a < b < c < d} \right.} \right)  \cr 
  & \quad  \Downarrow   \cr 
  & 0 < a = x_{\,1} \quad b = x_{\,2}  + 1\quad c = x_{\,3}  + 2\quad d = x_{\,4}  + 3  \cr 
  & \quad  \Downarrow   \cr 
  & N = 4!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {x_{\,1}  + x_{\,2}  + 1 + x_{\,3}  + 2 + x_{\,4}  + 3 = 21\quad
 \left| {\;0 < x_{\,1}  \le x_{\,2}  \le x_{\,3}  \le x_{\,4} } \right.} \right) =   \cr 
  &  = 4!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {x_{\,1}  + x_{\,2}  + x_{\,3}  + x_{\,4}  = 15\quad 
  \left| {\;0 < x_{\,1}  \le x_{\,2}  \le x_{\,3}  \le x_{\,4} } \right.} \right) =   \cr 
  &  = 4!\;{\rm N}{\rm .}\,{\rm of}\,{\rm partitions}\,{\rm of}\,15\;{\rm into}\,4{\rm parts} =   \cr 
  &  = 4!\;27 = 648 \cr} 
$$
which checks with direct computation.
To exemplify that, let's take a case with smaller values
$$
\eqalign{
  & N = {\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {a + b + c = 9\quad \left| {\;0 < a \ne b \ne c} \right.} \right)  \cr 
  & \quad  \Downarrow   \cr 
  & N = 3!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {a + b + c = 9\quad \left| {\;0 < a < b < c} \right.} \right)  \cr 
  & \quad  \Downarrow   \cr 
  & N = 3!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {x_{\,1}  + x_{\,2}  + 1 + x_{\,3}  + 2 = 9\quad \left| {\;0 < x_{\,1}  \le x_{\,2}  \le x_{\,3} } \right.} \right) =   \cr 
  &  = 3!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {x_{\,1}  + x_{\,2}  + x_{\,3}  = 6\quad \left| {\;0 < x_{\,1}  \le x_{\,2}  \le x_{\,3} } \right.} \right) =   \cr 
  &  = 3!\;{\rm N}{\rm .}\,{\rm of}\,{\rm partitions}\,{\rm of}\,6\;{\rm into}\,3\;{\rm parts} =   \cr 
  &  = 3!\;3 = 18 \cr} 
$$
and in fact
$$
\underbrace {\left[ {1,1,4} \right],\left[ {1,2,3} \right],\left[ {2,2,2} \right]}_{partit.\;x_{\,1}  \le x_{\,2}  \le x_{\,3} }\quad  \Rightarrow \quad
 \underbrace {\left[ {1,2,6} \right],\left[ {1,3,5} \right],\left[ {2,3,4} \right]}_{ordered\;0 < a < b < c}
$$
are the only ordered triplets of different  positive integers that sums to $9$.
Since they contain different integers, you can permute each of them
to obtain that the number of unordered triplets is $18$.
