All 1-50 solved.
Solved but extending meanings of operators: $28, 34, 38, 43, 44, 45, 46, 47, 49, 50$
$\frac{2}{0!}-2+0! = 1$
$2+\frac{0}{2}+0 = 2$
$2+\frac{0}{2}+0! = 3$
$2+0+2+0 = 4$
$2+0+2+0! = 5$
$2+0!+2+0! = 6$
$2^{0!+2} - 0! = 7$
$2^{0!+2} - 0 = 8$
$\frac{20}{2}-0! = 9$
$\frac{20}{2}-0 = 10$
$\frac{20}{2}+0! = 11$
$2 \cdot (0!+2)! + 0 = 12$
$2 \cdot (0!+2)! + 0! = 13$
$20 - (2+0!)! = 14$
$\frac{2+0!}{.2} + 0 = 15$
$2^{0!+2+0!} = 16$
$20 - (2+0!) = 17$
$20 - 2 + 0 = 18$
$20 - 2^0 = 19$
$20 - (2 \cdot 0) = 20$
$20 + 2 - 0! = 21$
$20 + 2 - 0 = 22$
$20 + 2 + 0! = 23$
$(2+0+2+0)! = 24$
$(2+0+2)! + 0! = 25$
$20 + (2+0!)! = 26$
$(2+0!)^{2+0!} = 27$
$\text{square}(2) \cdot (0!+(2+0!)!) = 28$ (using square
)
$\left \lfloor{\sqrt{20}}\right \rfloor! + \left \lfloor{\sqrt{20}}\right \rfloor = 28$ (using floor
)
$((2+0!)!+2-0!)!!! = 28$ (using Multifactorial, added by @Vepir)
$\frac{(2+0!)!}{.2} - 0! = 29$
$\frac{(2+0!)!}{.2} + 0 = 30$
$\frac{(2+0!)!}{.2} + 0! = 31$
$2^{(0!+2)!-0!} = 32$
$\sqrt[.2]{0+2} + 0! = 33$
$\sqrt[.2]{2} + 0! + 0! = 34$ (not in order)
$2\cdot(-0!+((2+0!)!)!!!) = 34$ (using unary minus
and Multifactorial, added by @Vepir in comments)
$(2+0!)!^2-0! = 35$
$\left(2+0!\right)!\cdot \left(2+0!\right)! = 36$
$((2+0!)!)^2+0! = 37$
$(20-0!) \cdot 2 = 38$ (not in order, added by @UnidentifiedX)
$20 + ((2+0!)!)!!! = 38$ (using Multifactorial)
$2\cdot (-0!+20)$ (using unary minus
, added by @Vepir in comments)
$20\cdot 2-0! = 39$
$20\cdot 2-0 = 40$
$20\cdot 2+0! = 41$
$2 \cdot (0!+20) = 42$ (added by @UnidentifiedX in comments)
$!(2+0!+2)-0!=43$ (using derangements, added by @Vepir in comments)
$!(2+0!+2)+0=44$ (using derangements, added by @Vepir in comments)
$!(2+0!+2)+0!=45$ (using derangements, added by @Vepir in comments)
$((2+0!)!)!! - 2 + 0 = 46$ (using Multifactorial)
$((2+0!)!)!! - !(2 + 0) = 47$ (using Multifactorial and derangements)
$2\cdot(0!+2+0!)! = 48$
$((2+0!)!+0!)^2 = 49$ (not in order, added by @JMP)
$((2+0!)!)!! + !(2 + 0) = 49$ (using Multifactorial and derangements)
$\frac{0!+0!}{.2^2} = 50$ (not in order, added by @JMP)
$((2+0!)!)!! + 2 + 0 = 50$ (using Multifactorial)