There are 2 blue and 3 red balls in a bag. If two balls are

Umm, 2C2 = (2!)/(2! x 0!)
and 0! = 1

Therefore, 2C2 = 2/(2 x 1) = 1

Now I will solve this question by using your approach. Your method is correct but need a little bit of adjustment in logic.

Question: Find Prob[ Blue >= 1]

This is kind of advance probability question. You can split Prob[ B >= 1] into 2 cases.

Prob[Blue >=1 ] = Prob[Blue = 1] + Prob[Blue = 2]

Prob[Blue=1] = (pick 1 blue) x (pick 1 red) / (pick 2 ball randomly)
= (2C1) x (3C1)/(5C2)
= (2 x 3) / (10) = 6/10

Prob[Blue=2] = (pick 2 blue) / (pick 2 ball randomly)
= (2C2) / (5C2)
= 1/10

Therefore; Prob[Blue >= 1] = 6/10 + 1/10 = 7/10


Or, you can use another method,which has already been shown by one of our friend above me, to solve this question.
Prob[Blue>=1] = 1 - Prob[Red=2]

Prob[Red = 2] = (3C2)/(5C2)
= 3/10

Prob[Blue>=1] = 1 - Prob[Red=2] = 1 - 3/10 = 7/10

I will solve it like this:

1 blue ball: 2C1 * 3C1 / 5C2 = 6/10 = 0.6
both balls are blue: 2C2/10 = 1/10 = 0.1

total p = 0.6 + 0.1 = 0.7