How many different arrangements are possible if 4 actors try out for 4 roles and no one will perfrom more?

Question

How many different arrangements are possible if 4 actors try out for 4 roles and no one will perfrom more than one role?

i want to know why the answer is 4x3x2 instead of 4x4

also, is this permutation or combination?

and when do i use 4x4?

Answer 1

4! is correct, and this is a permutation because actor A playing Romeo and actor B playing Juliet is different from the opposite.

You would use 4x4 if the actors/roles were independent. That is, an actor is not precluded from playing a role because he has already been cast in another.

For example, any of the 4 actors can play the first role, any of the 4 can play the second, any the third and any the fourth. 'Twould make a strange play.

Answer 2

Well, assume the first actor tries out for role 1.  Then how many ways can the other three roles be split among the 3 actors that are left?  

Let us start at the beginning:  how many ways can one role be arranged for one actor?  One.

How many ways for two roles and two actors?  Well, one actor takes the first, and the other actor takes the second.  So that is one.  Switch roles, that is another, so the answer is two.

How many ways for 3 actors and 3 roles?  Well, one actor takes the first, and there are two ways for the other actors to arrange the last two roles.  Let the first actor take another role, then there are two more arrangements for the other two roles.  (for a total of 4 so far), now let the first actor take the last role, then there are two more arrangements, for a total of 6 = 3*2*1.  You can see the pattern.

For n actors there are n! ways of arranging the roles.  Note n! (pronounced n factorial is n! = n*(n-1)*...*2*1