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$5$ tourists plan to attend Octoberfest. Each attends a location at random from the choices: Alpine, Bingeman, Concordia, Kitchener, Queens, Schwaben, Transylvania.

Q What is the probability that two attend one location and three attend another same location?

I get that there are $\binom{5}{2}$ possibilities to create 2 groups. There are $\binom{7}{2}$ ways to choose 2 places so shouldn't it be $\frac{\binom{7}{2}\binom{5}{2}}{7^5}$?

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    Properly speaking, that sentence should read: "Each attends a location *independently chosen* at random from the choices...." ;) – 2017-01-12
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    Your calculation is incorrect because the numerator assumes tourists are indistinguishable, and the denominator assumes they *are* distinguishable. – 2017-01-12

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You are almost right.

Hint: $\binom72$ is the number of ways to choose two different places from seven places, where the order of the two places is not important. Can you see (in the context of your problem) what is wrong with this?

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    I see, so it is 7*6 instead of that? – 2017-01-12
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    That's correct, well done. – 2017-01-12