<p>here it is:
There are exactly 4 actors which try out for the 4 parts in a play. If each actor can preform any one part and no one will perform more than one part, how many different assignments of actors are possible?</p>
<p>I said 16 but the answer was 24</p>
<p>4! = 4 x 3 x 2 x 1 = 24</p>
<p>I'm not really familiar with the whole idea of factorials, but just based on what the answer was, I knew it had to do with factorials. Study up on factorials, permutations, and combinations.</p>
<p>4 choices for the first spot, 3 for the second, 2 for the third, 1 for the fourth:</p>
<p>4x3x2x1 = 24</p>
<p>Yup it's 4!</p>
<p>wow this sat math is messing me up... thats grade 12 in canada where im from yet calc (not pre calc) is grade 11...</p>
<p>The permutation for 4 choices taken 4 at a time is
P(4,4) = 4!/(4-4)! = 4! / 0! = 4! = 24</p>
<p>lsfii2, I'm from Canada as well (mississauga, ontario), and had to really study ahead for some of the math stuff.</p>
<p>yea its weird i dread the number 12-14 geometry questions more than the function notation questions later in the section...</p>