# Permutation Question

• Aug 17th 2009, 04:25 PM
Jackson107
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???
• Aug 17th 2009, 04:45 PM
skeeter
Quote:

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???

$nPr = \frac{n!}{(n-r)!}$
• Aug 17th 2009, 04:48 PM
pickslides
$^nP_r = \frac{n!}{(n-r)!}$

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?
• Aug 17th 2009, 04:49 PM
Jackson107
i know that already but how do i complete the problem?
• Aug 17th 2009, 05:06 PM
Plato
Quote:

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

Use a calculator.
• Aug 17th 2009, 05:09 PM
pickslides
Quote:

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

$n! = n\times (n-1) \times (n-2) \times (n-3) \times \dots 3\times 2 \times 1$
• Aug 17th 2009, 07:53 PM
mr fantastic
Quote:

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.