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?

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 
$$\mbox{Number of ways to write } N \mbox{ as the sum of } k \mbox{ natural numbers } = {N - 1 \choose k-1 } = \frac{N-1!}{(k-1)! (N-k+1)!} $$
