In how many ways can you have at least two queens in a 6 card hand using a standard deck?

There are 4 queens in the deck and 48 non queens.

With two queens you want four non queens and the number of selections is 4C2×48C4=1167480

With three queens you want 3 non-queens and the number of selections is 4C3×48 C3=69184

With four queens you want 2 non-queens and the number of selections is 4C4×48C2=1128.

Summing gives 1137792, quite a lot!

Steve H answered your question literally. You just asked how many ways you can have at least 2 queens. I suspect you wanted to ask how many different 6-card hands there are with at least 2 queens.

You need to look at exactly how many queens you have

There is 4C4 = 1 way to have 4 queens, times 48C2 ways to fill out the other two cards.

There are 4C3 = 4 ways to have 3 queens, times 48C3 ways for the rest.

There are 4C2 = 6 ways to have 2 queens, times 48C4 ways for the rest.

Calculate those, multiply them out and add up the products for the answer.

You can have 2, 3 or 4 Queens from 4 and 4, 3 and 2 from the remaining 48 cards:

hands = C(4,2)C(48,4) + C(4,3)C(48,3) + C(4,4)C(48,2) = 1 237 792 <----