What is the probability that a randomly chosen $10$-card hand has exactly three three- of-a-kinds (and no four-of-a-kinds)? This is my attempt:
For the first three-of-a-kind:
There are ${13}\choose{1}$ options for the three cards alike and ${4}\choose{3}$ for the suits
For the second three-of-a-kind:
There are ${12}\choose{1}$ options for the three cards alike and ${4}\choose{3}$ for the suits
Similarly for the third: ${11}\choose{3}$ $\times$ ${4}\choose{3}$
Finally, for the remaining card, there are ${43}\choose{1}$ options for the remaining card and ${4}\choose{1}$ for the suit. 
So there are $\displaystyle\frac{C(13,1)\times C(4,3)\times C(12,1)\times C(4,3) \times C(11,1)\times C(4,3)\times C(43,1)\times C(4,1)}{C(52,10)}$ 
Can someone please check and verify? If this is wrong, can someone help me through this problem. 

We'll divide the number of successful hands by the total number of hands. 

The total number of hands is $$\binom{52}{10}$$

The number of successful hands is the number of ways to choose $3$ card values that will be repeated $3$ times, then $1$ card value that will occur once, then the excluded suit for each of the sets of $3$, and finally the suit of the singleton card:
$$\binom{13}{3}\binom{10}{1}\binom{4}{1}^3\binom{4}{1}$$

Our final probability is
$$\displaystyle\frac{\displaystyle\binom{13}{3} \cdot 10 \cdot 4^4\,\,}{\displaystyle\binom{52}{10}}$$
