Permutation allowing repetition

Question

Jeff is going to eat dinner out each day from Monday to Friday in a certain week, with each dinner being at one of his $15$ favorite restaurants.

$1$) How many possibilities are there for Jeff's schedule of dinners for that Monday to Friday if he does not want to eat at the same restaurant more than once?

$2$) How many possibilities are there for Jeff's schedule of dinners for that Monday to Friday if he is willing to eat at the same restaurant more than once, but not on two or more consecutive days?

I interpreted the first question as just a permutation question so I simply did $15$ $permutation$ $5$ to get $360360$ possibilities.

I first just wanted to know if that was indeed all there was to the problem.

Also, I am somewhat confused on how to do the second question as it seems like it requires a permutation but allows for repetition. But I am not sure how to accommodate for consecutive days and the same restaurant appearing more than once.

Any help would be highly appreciated!

Answer 1

It will be $15*(14)^4. $. $\\$ He can visit to one of the restaurant on Monday he has 15 option. $\\$ But for Tuesday there restaurant he visited on Monday he will not consider so can select one from rest 14 and Simillarly for next day 14 and so on till Friday. $\\$ Here in I am going with assumption he can go to same restaurant but not on alternate day .

Answer 2

You are right on the first one.

For the second, consider that there are $15$ restaurants Joe can choose from on Monday. On Tuesday, he can choose any restaurant but the one he chose on Monday, so he has $14$ choices. How nay choices has he on Wednesday?