Find the number of non-negative integral solutions of $x+y+z≤ 20$ Find the number of non-negative integral solutions of $x+y+z≤ 20$.

Conventional method :
x+y+z=20, then no. of ways= 22C2 (using beggar's method)
x+y+z=19, then no. of ways= 21C2
x+y+z=18, then no. of ways= 20C2
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x+y+z=0, then no. of ways= 2C2
Therefore total number of ways= 22C2 + 21C2+...3C2+ 2C2 = 22C2 + 21C2+...3C2+ 3C3 = 23C3
I was wondering if there is any other way to solve this question.
Let $w=20-x-y-z$. Then $w+x+y+z=20$ and $x,y,z,w$ are non-negative integers.
Sorry for the late answer. But I believe I found an even better way to solve this.
You can take a multinomial as $(a+a+a+1)^{20} $
The general term is coefficient × $a^{x+y+z} ×1^k$, where x+y+z+k=20
So note that it includes every case of x+y+z less than or equal to 20(non negative integral).
So the number of terms of the multinomial $^{23}C_3$ will give you the answer.
Simple!