How Many 5 Card Hands Have At Least 2 7s I am trying to find out how many 5 card hands have at least two 7s. I came up with the following solution: 
$${4 \choose 2}{48 \choose 3} + {4 \choose 3}{48 \choose 2} + {4 \choose 4}{48 \choose 1} = 108336$$
Is this correct or am I on the right track?

Yes, that is okay.   Count the suits for the sevens, for 2 to 4 sevens, and ways to select the remainder from the non-sevens, gives the count of 5 hand cards with at least 2 sevens. (Assuming a standard deck and no joker, as usual.)
$$\binom{4}{2}\dbinom{48}{3}+\binom{4}{3}\dbinom{48}{2}+\binom{4}{4}\dbinom{48}{1}$$

Alternatively, you could try complements, but it isn't any simpler in this case: $$\dbinom {52}{5}-\dbinom{48}5-\dbinom{48}4\dbinom{4}1$$
