What is the probability of getting at least 2 cards of the same suit, when dealt 3 cards from a standard deck

I dont know whether to use the choosing method, or a counting method..help!

"at least 2 cards of the same suit" is actually two different outcomes:
• exactly 2 cards of the same suit
or
•all 3 cards of the same suit
Note: the denominator of all possible selections is 52C3

Exactly 2 cards of the same suit: 4C1*13C2*3C1*13C1 = 4*78*3*13 = 12168
{In the above calculation, 4C1 is picking one suit of the four, 13C2 is picking two cards from that suit, 3C1 is picking one of the three remaining suits, 13C1 is picking one of the cards in that suit.}

Three cards of the same suit: 4C1*13C3 = 4*286 = 1144
{In the above calculation, 4C1 is picking one of the four suits, 13C3 is picking three cards of the 13 in that suit}

Total possible outcomes = 13312 divided by 52C3, which is 22100 = 13312/22100 = .60235

Two excellent answers. The first counts all the possible hands, which is good to be able to do.

But Mathmom's answer teaches something different. Analyze the problem and you can find a much simpler answer. It's very common in "at least x" type combination problems that her approach is the easier one.

P (at least 2 out of 3 cards have same suit)
= 1 − P(all cards different suit)
= 1 − 52/52 * 39/51 * 26/50
= 256/425
= 0.602352941