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= $^{22}C_2$ (using beggar's method)
$x+y+z=19$, then no. of ways= $^{21}C_2$
$x+y+z=18$, then no. of ways= $^{20}C_2$
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$x+y+z=0$, then no. of ways= $^{2}C_2$
Therefore total number of ways= $^{22}C_2$ + $^{21}C_2$ +... $^{3}C_2$+ $^{2}C_2 = ^{23}C_2$
I was wondering if there is any other way to solve this question.

If $x+y+z \leq 20$ Then the number of solutions will be the same as solving for the number of solutions to $w + x + y + z = 20$.
One approach to compute this is to look at the coefficient of $x^{20}$ in the generating function $f(x) = (1 + x + x^2 + ... + x^{20})^{4}$.
Use the identity $\displaystyle \sum_{n=0}^{k}x^n = \frac{1-x^{k+1}}{1-x}$:
$$[x^{n}]f(x) = [x^{n}](1 + x + x^2 + ... + x^{20})^{4}$$
$$[x^{n}]f(x) = [x^{n}]\left(\dfrac{1-x^{21}}{1-x}\right)^{4}$$
$$[x^{n}]f(x) = [x^{n}]\dfrac{x^{84} - 4 x^{63} + 6 x^{42} - 4 x^{21} + 1}{(1-x)^4}$$
Use the identity $\displaystyle \frac{1}{(1-x)^{m+1}} = \sum_{n=0}^{\infty}\binom{n+m}{m}x^n$:
$$[x^{n}]f(x) = [x^{n}](x^{84} - 4 x^{63} + 6 x^{42} - 4 x^{21} + 1)\sum_{n=0}^{\infty}\binom{n+3}{3}x^n$$
$$[x^{n}]f(x) = [x^{n}]\sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+84} - 4 \sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+63} + 6 \sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+42} \\- 4 \sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+21} + \sum_{n=0}^{\infty}\binom{n+3}{3}x^n$$
Shift indices:
$$[x^{n}]f(x) = [x^{n}]\sum_{n=84}^{\infty}\binom{n-81}{3}x^{n} - 4 \sum_{n=63}^{\infty}\binom{n-60}{3}x^{n} + 6 \sum_{n=42}^{\infty}\binom{n-39}{3}x^{n} \\- 4 \sum_{n=21}^{\infty}\binom{n-18}{3}x^{n} + \sum_{n=0}^{\infty}\binom{n+3}{3}x^n$$
Now we take the coefficient of $x^{20}$:
$$[x^{20}]f(x) = \binom{20+3}{3} = \binom{23}{3}$$
Note that most of the binomial sums dropped out because their starting indices were greater than $20$ (and therefore there were no $x^{20}$ terms present in any of them), but you could easily recompute the number of ways for any given boundary (not just $20$) and take the relevant binomial coefficients.
For example the number of ways for $w+x+y+z = 50$ where $0 \leq w, x, y, z \leq 20$ would be $[x^{50}]f(x) =  6\binom{50-39}{3}- 4 \binom{50-18}{3} + \binom{50+3}{3} = 4576$
