ok everybody i have algebra 2 and im doing permutations and i have a question like this on my paper:
4P3
----=
8P3
can you guys help me out please???
$\displaystyle ^nP_r = \frac{n!}{(n-r)!}$
Consider
$\displaystyle ^4P_3 = \frac{4!}{(4-3)!}$ and $\displaystyle ^8P_3 = \frac{8!}{(8-3)!}$
Therefore
$\displaystyle \frac{^4P_3 }{^8P_3}= \frac{\frac{4!}{(4-3)!}}{\frac{8!}{(8-3)!}} = \frac{\frac{4!}{1!}}{\frac{8!}{5!}} = \frac{4!}{1!}\times\frac{5!}{8!}$
Can you finish it from here?
If you knew all that already it would have been good for you to have said so in the OP so that people didn't end up wasting their time explaining it. And it wouldn't have hurt to Thank those people for their effort.
To complete the problem you substitute the fact that 8! = 8x7x6x5!, cancel the common factor of 5! and then do some basic arithmetic.