# Probability/counting problem.

In an experiment E, nine people are asked their birth MONTH, and the nine responses are then written down. All outcomes are equally likely.

Find the probability of the event B = {NO TWO people are born in the same month.}

I know the sample space is 12^9. But i cant figure out how to get the size of B.

I know it is (12)sub9 or 12!/3! because i have the answer key. But I can't figure out why. The answer key doesn't explain. Why is the answer not just 12 choose 9? I feel the answer is 12 choose 9 because you are picking 1 month for each person, without replacement to deal with the restriction, and there are 9 people.

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12!/3! is huge. I'm pretty sure that can't be the solution. – Newb Oct 28 '13 at 5:05
That isn't the probability, it's the size of B according to the answer key. – mashedtatoes Oct 28 '13 at 5:06
If the sample space is $12^9$, that means you're treating the nine responses as distinguishable (e.g., maybe everyone writes their name as well). So "January and eight Februaries" is a different outcome than "four Februaries, a January, and four more Februaries". In that case, the number of outcomes where no two people share a month should be ${{12}\choose{9}}$ (choose the months) times $9!$ (choose the order), or $12! / 3!$. – mjqxxxx Oct 28 '13 at 5:11
Now that makes perfect sense! I didn't think about the ordering. Thank you! – mashedtatoes Oct 28 '13 at 6:12

$$\frac{12}{12} \times\frac{11}{12} \times \ldots \times \frac{4}{12} \approx .015$$