Question:problem from prep book:

exactly 4 actors try out for the 4 parts in a play. if each actor can perform any one part and no one will perform more than one part, how many different assignments of actors are possible?

the answer is 24. but how did the book arrive at that answer?

Possibility problems like this usually unvolve factorials (4! = 4x3x2x1).

Think of it this way. For the first role, there are four possible actors to choose from, meaning 4 possibilities. After that role is filled, there are only 3 remaining choices for the next role. After that role is filled, there are 2 remaining actors to choose from for the third role. And after all these roles are filled, the last remaining actor must fill the fourth role.

So you multiply these possibilities by each other to get the total number of possible combinations: 4x3x2x1 = 24
4x3x2x1=24

lets say each actor is A B C D

You have to solve for how many different orders in which you can put them in. Well you could go and list off each one,

for exp
ABCD
ABDC
ACDB

etc.

Its difficult to explain.
Just look up some basic info on probability.

This article contents is post by this website user, EduQnA.com doesn't promise its accuracy.