- #1
MiniTank
- 62
- 0
how would you solve for r in this situation:
P(5, r) = 20
Now I understand you can't use trial and error to find that r=2 but is there any other way using an algebraic technique. Itried something like this..
[tex] \frac{5!}{(5-r)!}=20[/tex]
then
[tex]\frac{5\times4\times3\times2\times1}{20}=(5-r)![/tex]
[tex]6=(5-r)![/tex]
now i don't know what to do next
P(5, r) = 20
Now I understand you can't use trial and error to find that r=2 but is there any other way using an algebraic technique. Itried something like this..
[tex] \frac{5!}{(5-r)!}=20[/tex]
then
[tex]\frac{5\times4\times3\times2\times1}{20}=(5-r)![/tex]
[tex]6=(5-r)![/tex]
now i don't know what to do next