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?
 A: Hint:
Assuming that numbers are able to be repeated.
Place $21$ identical stars in a row.  Simultaneously place $3$ dividers in spaces between the stars with at most one divider per space.  Now, interpret the resulting configuration as a desired arrangement.
Convince yourself that every possible arrangement of stars and bars uniquely corresponds to one of your desired solutions and vice versa.  Now, count how many ways you can arrange the stars and bars.
E.g.
$\star\star\star|\star\star\star\star\star|\star\star\star\star\star\star|\star\star\star\star\star\star\star$ corresponds to $(3,5,6,7)$
This proof-technique is referred to as stars-and-bars.

If numbers aren't able to be repeated, the above technique can still be used in conjunction with inclusion-exclusion to account for getting rid of invalid cases.
A: $$
\eqalign{
  & N = {\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {a + b + c + d = 21\quad \left| {\;0 < a \ne b \ne c \ne d} \right.} \right)  \cr 
  & \quad  \Downarrow   \cr 
  & N = 4!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {a + b + c + d = 21\quad \left| {\;0 < a < b < c < d} \right.} \right)  \cr 
  & \quad  \Downarrow   \cr 
  & 0 < a = x_{\,1} \quad b = x_{\,2}  + 1\quad c = x_{\,3}  + 2\quad d = x_{\,4}  + 3  \cr 
  & \quad  \Downarrow   \cr 
  & N = 4!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {x_{\,1}  + x_{\,2}  + 1 + x_{\,3}  + 2 + x_{\,4}  + 3 = 21\quad
 \left| {\;0 < x_{\,1}  \le x_{\,2}  \le x_{\,3}  \le x_{\,4} } \right.} \right) =   \cr 
  &  = 4!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {x_{\,1}  + x_{\,2}  + x_{\,3}  + x_{\,4}  = 15\quad 
  \left| {\;0 < x_{\,1}  \le x_{\,2}  \le x_{\,3}  \le x_{\,4} } \right.} \right) =   \cr 
  &  = 4!\;{\rm N}{\rm .}\,{\rm of}\,{\rm partitions}\,{\rm of}\,15\;{\rm into}\,4{\rm parts} =   \cr 
  &  = 4!\;27 = 648 \cr} 
$$
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 $18$.
A: This is the answer to "How many ways are there to distribute 21 candies to 4 children, where each child gets at least 1 candy?"
To do this, lay out 21 candies in a row:
$$ \underbrace{\bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet}_{21  \,\,\, Candies} $$
To produce the sum, say $3+5+6+7=21$, we can separate the candies with dividers as such:
$$ \bullet \bullet \bullet | \bullet \bullet \bullet \bullet \bullet | \bullet \bullet \bullet \bullet \bullet \bullet | \bullet \bullet \bullet \bullet \bullet \bullet \bullet $$
Notice there are 20 choices for the location of the dividers, and we must pick 3.  The question then becomes,

How many ways are there to choose 3 spots out of 20 to place dividers?

The answer is $$ {21 - 1 \choose 4-1 } = {20 \choose 3 } = \frac{20!}{3! 17!}$$
In general, the answer is 
$$\mbox{Number of ways to write } N \mbox{ as the sum of } k \mbox{ natural numbers } = {N - 1 \choose k-1 } = \frac{N-1!}{(k-1)! (N-k+1)!} $$
