There are 55 ways to choose the first Republican, 54 ways to choose the second Republican, and 53 ways to choose the third Republican; however, we must divide by $3!$ because order does not matter.  So the number of ways to choose Republicans is $\dfrac{55 \times 54 \times 53}{3!} = 26,\!235$. There are 45 ways to choose the first Democrat and 44 ways to choose the second Democrat, but we must divide by $2!$ because order does not matter.  So the number of ways to choose the Democrats is $\dfrac{45 \times 44}{2!} = 990$.  So there are $26,\!235 \times 990 = \boxed{25,\!972,\!650}$ ways to choose a committee.
The Senate has 100 members, 55 Republicans and 45 Democrats.  In how many ways can I choose a 5-person committee consisting of 3 Republicans and 2 Democrats?
{'answer': '25,\\!972,\\!650', 'subject': 'Counting & Probability', 'level': 3, 'unique_id': 'test/counting_and_probability/495.json'}