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A bakery produces six different kinds of pastry. If the different kinds of selections of 20 pastries are equally likely, what is the probability that a selection contains exactly three eclairs?

In discrete the problem is solved as follows, (17+5-1)C(17)=5985 is the number of ways to get exactly 3 eclairs. The total number of ways to choose is (20+6-1)C(20)=53130. So the probability of getting exactly 3 eclairs is 5985/53130=.113. The book has the same answer.

But in statistics I'd solve the problem as a binomial. The chance of selecting an eclair would be viewed as a success (1/6 chance). So just plugging this into my calculator for a quick answer I do bipdf(20,1/6,3)=.238, & I get the same thing when I use the formula by hand.

So why are these values different? The question is from the discrete book, so is my statistics wrong?