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Six people sat down along one side of a banquet table completely ignoring their name cards. In how many ways could this have been done so that no person was seated where his/her name card was placed?

I think the solution is 5^6 because each person can go to other 5 chairs that is not seated... However the answer is 265. Can someone explain this to me?

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    $5^6$ would allow for things such as everyone who was supposed to go through seats $2$ through $6$ to all try to sit in seat $1$. I don't know if you've ever tried to have five people all sit in the same seat simultaneously but it doesn't work well. The term for the type of permutation you are being asked to investigate is called a [derangement](https://en.wikipedia.org/wiki/Derangement). There are many pages already on this site dedicated to proving the formulae associated with derangements. I suggest you search with this term a bit longer and come back if you still can't figure it out. – 2017-02-03
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    I don't know, sometimes it works really *really* well. Don't knock it till you've tried it. – 2017-02-04

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To remove this from the queue:

The answer is $!6=265$ where $!n$ denotes the subfactorial function which counts the number of derangements on a set of $n$ elements. A derangement is a permutation with no fixed points, as is the case with your example of people sitting around the table in any place but their assigned seat.

Various methods exist for counting derangements, perhaps the most accessible in your case is:

$$!6=6!\sum\limits_{i=0}^6\frac{(-1)^i}{i!}$$

which can be derived via inclusion-exclusion on the events that seat $i$ has person $i$ sitting in it.