Answer
$52$ ways.
Work Step by Step
First, let's find $N(3~twos)$. The order in which the $twos$ are select does not matter and no $two$ can be selected more than once.
It is a combination of 4 $twos$ taken 3 at a time:
$_4C_3=\frac{4!}{3!(4-3)!}=\frac{4!}{3!\times1!}=\frac{4\times3\times2\times1}{3\times2\times1\times1}=4$
But, $N(3~twos)=N(3~threes)=N(3~fours)=N(3~fives)=N(3~sixs)=N(3~sevens)=N(3~eights)=N(3~nines)=N(3~tens)=N(3~jacks)=N(3~queens)=N(3~kings)=N(3~aces)=4$
$N(3~of~the~same)=N(3~twos~or~3~threes~or~3~fours~or~3~fives~or~3~sixs~or~3~sevens~or~3~eights~or~3~nines~or~3~tens~or~3~jacks~or~3~queens~or~3~kings~or~3~aces)$ $=N(3~twos)+N(3~threes)+N(3~fours)+N(3~fives)+N(3~sixs)+N(3~sevens)+N(3~eights)+N(3~nines)+N(3~tens)+N(3~jacks)+N(3~queens)+N(3~kings)+N(3~aces)=13\times4=52$
