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.

  • $\begingroup$ 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? $\endgroup$
    – Glen_b
    Sep 15, 2014 at 2:25

1 Answer 1


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.


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