How's It Possible?

The required number of ways is $\begin{aligned} \sum_{r=0}^{20} \binom{20}{r} r! &= \sum_{r=0}^{20} \binom{20}{r} \int_{0}^{\infty} e^{-x} x^r \mathrm{d}x\\ &= \int_{0}^{\infty} \sum_{r=0}^{20} \binom{20}{r} e^{-x} x^r \mathrm{d}x\\ &= \int_{0}^{\infty} e^{-x} (1+x)^{20} \mathrm{d}x\\ &=\boxed{e \cdot \Gamma(21,1)}\end{aligned}$

Note : $\Gamma (a,x)$ is the upper incomplete gamma function.