# Thread: Permutation Question

1. ## Permutation Question

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???

2. Originally Posted by Jackson107
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???
$nPr = \frac{n!}{(n-r)!}$

3. $^nP_r = \frac{n!}{(n-r)!}$

Consider

$^4P_3 = \frac{4!}{(4-3)!}$ and $^8P_3 = \frac{8!}{(8-3)!}$

Therefore

$\frac{^4P_3 }{^8P_3}= \frac{\frac{4!}{(4-3)!}}{\frac{8!}{(8-3)!}} = \frac{\frac{4!}{1!}}{\frac{8!}{5!}} = \frac{4!}{1!}\times\frac{5!}{8!}$

Can you finish it from here?

4. i know that already but how do i complete the problem?

5. Originally Posted by Jackson107
i know that already but how do i complete the problem?
Use a calculator.

6. Originally Posted by Jackson107
i know that already but how do i complete the problem?
If you want to avoid using a calculator as Plato has suggested use my previous post and the fact that

$n! = n\times (n-1) \times (n-2) \times (n-3) \times \dots 3\times 2 \times 1$

7. Originally Posted by Jackson107
i know that already but how do i complete the problem?
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.