# drawing balls from an urn

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.

• As routine bookwork, this should carry the self-study tag. Please read the self-study tag wiki, add the tag and edit your question as suggested there. – Glen_b Sep 15 '14 at 2:13
• @Glen_b how is that? – statsnewb Sep 15 '14 at 2:18
• A definite improvement. Is there anything in particular making you unsure of your answer? Can you explain the reasoning by which you arrive at it? – Glen_b Sep 15 '14 at 2:25

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.