Suppose there are 7 golf balls in an urn and you draw balls in 3 stages without replacement. First you draw 2, then 3, and then the remaining 2. How many ways are there to do this?
I believe the answer would be ${7\choose 2}{5\choose 3}$ but I am making this question up so would like to verify the answer.
Your answer appears correct.
In particular, the number of ways will be the number of ways of choosing groups of 2,3 and 2 from a set of 7.
This is the multinomial coefficient ${7 \choose 2, 3, 2} = \frac{7!}{2!\, 3! 2!}$.
As you state in your question, its value is equal to ${7\choose 2}{5\choose 3}$, which you can arrive at via a number of different routes.
