Probability of having at least one pair by drawing 4 shoes from 12 pairs.
Calculate $1$ minus the probability of the complementary event:
The number of ways to choose $4$ out of $24$ shoes is:
- Choose the $1$st shoe out of $24$ shoes
- Choose the $2$nd shoe out of $23$ shoes
- Choose the $3$rd shoe out of $22$ shoes
- Choose the $4$th shoe out of $21$ shoes
The number of ways to choose $4$ out of $24$ shoes with no pairs is:
- Choose the $1$st shoe out of $24$ shoes
- Choose the $2$nd shoe out of $22$ shoes
- Choose the $3$rd shoe out of $20$ shoes
- Choose the $4$th shoe out of $18$ shoes
So the probability of choosing $4$ out of $24$ shoes with at least one pair is:
$$1-\frac{24\cdot22\cdot20\cdot18}{24\cdot23\cdot22\cdot21}$$
Please note that I've essentially taken into account the order of the shoes.
If I chose not to take it into account, then I would need to divide each result by $4!$.
But since this factor appears in both the numerator and the denominator, I can ignore it.
Your first method may double-count the possibility of getting two pairs: when you "chose any $2$ from the rest in ${22 \choose 2}$ ways", you may be choosing another pair and these two pairs are also counted when chosen in the other order.
In your second method of looking at $1-$ the probability of choosing from different pairs, a similar method is to say that each time you choose a shoe its pair becomes undesirable, making the result $$1-\frac{24}{24}\times \frac{22}{23}\times \frac{20}{22}\times \frac{18}{21} =\frac{41}{161}.$$
Just to build on Henry's answer, suppose we put the pairs of shoes in order, 1 through 12. By your method, you are over counting picking pair 1 and another pair 11 times. Similarly, you are over counting picking pair 2 with another pair (other than pair 1) 10 times. Continuing on, you are over counting by precisely 11th triangular number, $T_{11} = 66$. Correcting for this, $$ \frac{12\cdot {22 \choose 2}-66}{24\choose 4} = \frac{41}{161} $$ as desired.