Need help with probability!
(Free Response)
14. 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?
Since there are 4 actors and 4 parts I put 16. But the answer came out to be 24…not sure why. 
Questions like these make end up with a 760-780 instead of an 800. :c
Pls halp.
Each of the 4 actors may fill the first role, leaving 3 actors to fill the second role, leaving 2 actors to fill the third role, leaving one actor to fill the last role…
432*1 = 24
4! = 24 ways to assign roles, as Frigidcold said. (n! = n factorial, the number of ways to arrange n different objects in a line)
revise permutations for this. nPn = n!, 4P4 = 4! = 4 x 3 x 2 x 1 = 24
OP, the problem with your original answer is that it assumed that one person could have all 4 of the parts, or 3 of the 4, and so on.
Here’s the logic in the correct answer.
Any of the 4 could get part A. OK, so that part is taken, as is that actor.
That leaves 3 remaining actors for part B.
And 2 for part C.
And 1 for part D.
Since we’re auditioning for all 4 (“And” as opposed to “Or”), we multiply: 4x3x2x1 = 24.