I am trying to solve the following question.
How many different binary sequence pairs are there for a 4 digit and 5 digit sequence, where the first 4 digit sequence has 2 0's and 2 1's and the second 5 digit sequence has 2 0's and 3 1's.
My thought process is as follows,
For the first sequence there is $\binom{4}{2}$ and for the second sequence there is $\binom{5}{2}$.
So I must multiply these together no? Which is $$\frac{4!\cdot 5!}{(2)!(3)!\cdot 2! \cdot 2!} = 60$$
At this point I am not sure if I have overcounted, is this figure counting the distinct possibilities where the 4 digit sequence is first and the 5 digit sequence is second and vice versa. So should I divide by 2 in this case? I am not sure?
No, you've done it correctly. You haven't over-counted, because there is no way to confound a 4-bit sequence and a 5-bit sequence. If you had, say two 5-bit sequences, then you'd have to worry about double counting, if the order didn't matter.
Choosing A and then B is the same as choosing B then A, but only if both choices are possible. If A is to be a 4-bit sequence and B a 5-bit sequence, there is no possibility of choosing B and then A if you are choosing the 4-bit sequence first.