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There are 5 tourists going to London. There are 7 places to visit in London. What is the probability that two attend one (same) location and 3 attend another (same) location.

So this is asking what is the probability that the 2 groups of people attend different places, but everyone from the groups is there. ex. $AB | C D E$ for the groups.

There are $\binom{5}{2}$ ways to choose 2 groups. There are $7^5$ total choices considering each person has seven choices.

There are $\binom{7}{2}$ ways to pick 2 locations from the 7.

Thus, $P = \frac{\binom{5}{2} \binom{7}{2}}{7^5}$

But the answer the book has is

$$P = \frac{7*6 \binom{5}{2}}{7^5}$$

Why are the locations suddenly distinguishable?

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You picked two locations, but you also have to decide which group goes to which location - that's the extra factor of two in the book answer (since $2\binom 72 = 7\cdot 6$).

Otherwise - good analysis.