Permutation with $n$ repetitions of each element? Here's the problem:
A man has dinner 5 days a week, Monday to Friday, choosing from 10 restaurants. Thus, there are $$10\times9\times8\times7\times6$$
ways that he can have dinner if no repetitions are allowed.
Now, what if he can eat at each restaurant at most twice?
My first idea is that there are $\binom{20}{5}$ ways to pick the restaurants, where each of the 20 is a chance for the restaurant to get picked. Then if you want to permute them, you have $\binom{20}{5} \times 5!$ ways the man can eat dinner that week, but this is overcounting since for example the cases of eating at restaurant #1 on Monday and Wednesday, and at restaurant 1 on Wednesday and Monday are being counted as separate cases when they should be the same.
Is there a better approach to solve this problem?

three methods all give the same answer 91440
method 1
$10^5$ - number of ways to go to any restaurants any time
$10 * C^3_5 * 9 * 9 + 10 * C^4_5 * 9 + 10$ - number of ways to choose restaurants such that there is one restaurant chosen at least 3 times.
The answer is then $10^5 - 10 * C^3_5 * 9 * 9 - 10 * C^4_5 * 9 - 10 = 91440$
method 2
$10 * 9 * 8 * 7 * 6$ - choosing restaurants each at most one time
$C^2_5 * 10 * 9 * 8 * 7$ - choosing restaurants such that there is exactly one visited twice, and three others one time
$(C^2_5 * 10 * 9 * C^2_3 * 8) / 2$ - choosing restaurants such that there are two restaurants both visited twice and, one more restaurant visited once.
thus, answer is $10 * 9 * 8 * 7 * 6 + C^2_5 * 10 * 9 * 8 * 7 + (C^2_5 * 10 * 9 * C^2_3 * 8) / 2 = 91440$
method 3
numeric gives 91440: https://paste.ofcode.org/VyS2NMg6b6VyNNBBAccahw
