Probability that at least one of 2 balls taken randomly from a pile of 2 red and 3 black is red I get 2 different answers, depending how I approach this, and I need help to see why the error arises.

One solution is to calculate unfavorable combinations probability and substract from 1:
$$1-\frac{C_3^2}{C_5^2}=\frac{7}{10}$$
The other solution is when calculating favorable combinations, to first choose one of the 2 reds, and then choose one of the remaining 4:
$$\frac{C_2^1 \times C_4^1}{C_5^2} = \frac{8}{10}$$
Which is obviously $\neq \frac{7}{10}$.
The first solution is correct. The second solution makes the mistake of counting twice the scenario where both red balls are picked, which has a $\frac1{10}$ chance of occurring. Subtracting this from $\frac8{10}$ yields the correct answer of $\frac7{10}$.
Pick first red ball R1 and you have 4 possible combinations with the other balls: R1R2, R1B1, R1B2, R1B3
Now pick the second red ball R2 and you have 3 possible combinations left without repeating previous combinations (R2R1): R2B1, R2B2, R2B3
Continue with the first black ball B1: B1B2, B1B3
And the second black ball B2: B2B3
That's it, now you have 7 combinations with at least 1 red ball and 3 combinations with only black balls: 7/10