combinatorics - Incorrect combinatorial argument- 5 card hand with at least 3 red cards

How many 5 card hands can be made with at least three red cards? Of course, we're using a standard deck of 52. I know how to answer this, but I frequently see this argument, producing a different answer. I know it's wrong but I can't explain exactly what's wrong with it?

"First, there are $_{52}C_3$ of choosing three red cards. Since the other two cards can be black or red, we can choose them from any of the 49 unused cards. I.e. $_{49}C_2$ ways. So the final answer should be $_{52}C_3$ x $_{49}C_2$"

Answer

This counting argument is incorrect because it will multiple-count configurations for which more than three cards are red; e.g., $$\{10\color{red}\heartsuit, J\color{red}\heartsuit, Q\color{red}\heartsuit, K\color{red}\heartsuit, A\color{red}\heartsuit\} = \{\{10\color{red}\heartsuit, J\color{red}\heartsuit, Q\color{red}\heartsuit \}, \{K\color{red}\heartsuit, A\color{red}\heartsuit\}\} = \{\{10\color{red}\heartsuit, J\color{red}\heartsuit\}, \{ Q\color{red}\heartsuit, K\color{red}\heartsuit, A\color{red}\heartsuit\}\}$$ but under such an enumeration scheme, the middle and right hand expressions are considered distinct.