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

Answer 1

There are 4 different actors who could take the first role, 3 remaining who could take the second role, 2 remaining who could take the third role, and 1 left for the last role.  So the answer is 4! = 4*3*2*1 = 24.

The reason it's not 4*4 is because we're trying to count all the possible CASTS we could have for the an act, not all of the different possible acting positions.  If you were to make a 4x4 grid with the labels "Actor A, Actor B, Actor C, Actor D" across the top and "Role #1, Role #2, Role #3, Role #4" along the side, it's true that you'd get 16 positions listed: A1, A2, A3, A4, B1, B2, B3, B4, etc.  However, there's only 24 different ways you can pick a unique set of 4 of them.

Answer 2

There are 4 ways to choose the actor for the first role. Once that role is chosen, there are 3 actors left. Thus, for each of the 4 choices for the first role, there are 3 ways to fill the second role. That makes 4*3 = 12 ways so far. Likewise, now there are 2 actors left from which to choose the third role, making 4*3*2 = 24 ways. There is now only one actor left to choose the last role, and 4*3*2*1 = 24. This is the definition of factorial: 4! = 24.

It is not 4*4 because that is the number of ways 4 actors can be assigned to TWO roles if each can perform more than one role. There are 4 choices for the first role times four for the second making 16 ways.

Answer 3

The first actor has 4 different choices

The 2nd actor only has 3

The 3rd actor has 2

The last actor has 1

Therefore it is 4x3x2x1 (written as 4!)