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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
up vote 1 down vote accepted

Another way to solve this problem is to think sequentially. What's the probability of having all different birthmonths with just 1 person? 1 (duh). What if we add another person? What's the probability that this new person has a different birthmonth from everybody else? 11/12. If we keep going we see the answer is:

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

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