Make 1-100 using 2 0 2 2

Question

You may use the following:

• Addition [e.g. 2+2]
• Subtraction [e.g. 2-2]
• Multiplication [e.g. 2x2]
• Division [e.g. 2/2]
• Decimal point [e.g. 2+.2]
• Exponents [e.g. 2²] {tetration is NOT allowed}
• Square roots [e.g. sqrt(2)] {infinite roots sqrt(sqrt(sqrt…sqrt(2))))=1 is NOT allowed}
• Arbitrary roots [e.g. 0.2th root of 2]
• Factorial [e.g. (2+2)!]
• Double factorial [e.g. (2+2)!!]
• Parentheses [e.g. 2/(2+2)]
• Concatenation [e.g. 22+20] {2[0!]=21 is NOT allowed}
• Permutations [e.g. (2+2)P2]
• Combinations [e.g. (2+2)C2]

I’ve got all the numbers from 1-100, except for 54, 67, 68, 69, 79, 82, 83, 84, 86, 87, 93, 97

Answer 1

All remaining 12 numbers you listed are doable, only if you allow arbitrary "$.$" and "$!$" uses.

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Allowing arbitrary decimal point $.\square$, then 3 of remaining 12 numbers are:

$$\begin{array}{}54 &=& \sqrt{\text{}^{.(0!)}\sqrt[]{2}} + 22\\68 &=& (\sqrt{\text{}^{.(0!)}\sqrt[]{2}}+2) \times 2\\82 &=& ((2 \times 2)!! + .2) / .(0!)\end{array}$$

Additionally allowing repeated-decimals $.\dot{\square}$, then 7 of remaining 9 are:

$$\begin{array}{}67 &=& \left(\left(\sqrt{(.2) ^ {-2}}\right)!! - (.\dot{(0!)})\right) / (.\dot{2}) \\79 &=& (.\dot{(0!)}) ^ {-2} - 2\\83 &=& (.\dot{(0!)}) ^ {-2} + 2 \\84 &=& (.\dot{(0!)}) ^ {-2} + \sqrt{2 / (.\dot{2})} \\86 &=& (.\dot{(0!)}) ^ {-2} + \sqrt{(.2) ^ {-2}} \\87 &=& (.\dot{(0!)}) ^ {-2} + \left(\sqrt{2/(.\dot{2})}\right)! \\97 &=& \left(\left(\sqrt{2 / (.\dot{2})}\right)!\right)!! \times 2 + 0!\end{array}$$

Additionally allowing subfactorial $!\square$, then 2 of remaining 2 are:

$$\begin{array}{}69 &=& \left((0! / .2)!! + \sqrt{.\dot{(!2)}}\right) / (.\dot{2}) \\93 &=& (\text{}^{.2}\sqrt{2}-0!)/ \sqrt{.(\dot{!2})}\end{array}$$

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I wrote a python program to solve this.