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There are 3 basic categories here, as there has to be at least one of one number, and two of the other two numbers. So we have:

$\text{01122}$
$\text{10022}$
$\text{20011}$

So I take each these and permutate. Which gives me $3 \times 5!$ which is wrong. The answer is $90$. I know it is wrong because $3 \times 5! > 3^5$ which shouldn't be possible. I can't find any intuitive reason why my answer is wrong.
How did I approach it wrong?

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    $\begingroup$ There aren't $5!$ permutations of, say, $01122$. $\endgroup$ – lulu Jun 20 '19 at 0:33
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There are some repeated bits (like $1$'s and $2$'s in the first category). So you can't permute them with $5!$ since swapping $1$'s or $2$'s in the first category doesn't change the string, similar in the second and the third category. Therefore, your answer should be $$3\cdot\frac{5!}{2!2!} = 90$$

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