Standard proof of permutations of $n$ people standing in line

Question

I want to prove the statement: For $n$ people the number of permutations is $n!$. How to prove or justify this statement ?

I think the easiest way is to use induction.

If we have a line of $1$ people the number of permutations is 1.

For $n$ people the number of permutations is $n!$. Since we have the start we only need to prove that it holds for $n+1$.

$ (n+1)!=(n+1) n! = (n+1) $ "times we can reorder a line of n people" = times we can reorder a line of $n+1$ people.

Is this a good argument? Is there a more intuitive or elegant way to show it?

Answer 1

You have a row of $n$ places for $n$ people

• For the first place you can put any of $n$ people
• For the second place you can put any of $n-1$ people (since you have already put one in the first place)
• ...
• For the $n$th place you can put just 1 person (since you have already put other $n-1$ people to previous $n-1$ places)

So, using multiplication rule there are in total: $n \cdot (n-1) \dots 2 \cdot 1 = n!$ ways to place $n$ people