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???

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- Aug 17th 2009, 04:25 PMJackson107Permutation 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 PMskeeter
- Aug 17th 2009, 04:48 PMpickslides
$\displaystyle ^nP_r = \frac{n!}{(n-r)!}$

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? - Aug 17th 2009, 04:49 PMJackson107
i know that already but how do i complete the problem?

- Aug 17th 2009, 05:06 PMPlato
- Aug 17th 2009, 05:09 PMpickslides
- Aug 17th 2009, 07:53 PMmr fantastic
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.