In how many ways can ww choose $10$ cards so there are $3$ exact matches? Suppose there are $20$ cards- $10$ reds numbered $1,2,\cdots,10$, and $10$ blues numbered $1,2, \cdots, 10$. In how many ways can 10 cards be picked to so that there are EXACTLY 3 MATCHES- where a match means a red card and blue card have the same number. 
What I did:
Number of ways to get 3 or more matches - Number of ways to get 4 matches or more, that is:
$\dbinom{10}{3} \cdot \dbinom{14}{4} - \dbinom{10}{4}\cdot \dbinom{12}{2}$
Can someone point out what I did wrong?

There are $\binom{10}{3}$ ways to pick the cards that will make up the three matching pairs. That is a component of your answer, so I will assume that part is clear to you. Now we count the number of ways to pick the remaining $4$ cards. 
There remain $7$ "couples." We pick $4$ of these, and for each couple we choose the colour that will be used. That can be done in $\binom{7}{4}2^4$ ways, for a total of $\binom{10}{3}\binom{7}{4}2^4$.
Remark: About your answer, the issue is about the count of "bad" choices, where there are $4$ or more matches. This number is not $\binom{10}{4}\binom{12}{2}$. That expression "double-counts" the $5$ matches situations. For the $\binom{10}{4}$ part counts, among others, the situations where we picked each of the couples $1,2,3,4$, while the $\binom{12}{2}$ counts among others the situations where we picked couple $5$. But $\binom{10}{4}$ also includes picking couples $1,2,3,5$, and $\binom{12}{2}$ counts among others picking couple $4$. Thus picking the $5$ couples has been double-counted, indeed multiple-counted. 
