and where does the 5 outside of the (nP_5) come from??
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OpenStudy (asnaseer):
it is just a multiplying factor, e.g.:\[Q\times_nP_r=\frac{Q\times n!}{(n-r)!}\]
OpenStudy (thatonegirl_):
so wouldnt r be 25 then?
OpenStudy (thatonegirl_):
\[n!\div(n-r)!=5n!\div(5n-30)!\]
OpenStudy (asnaseer):
lets take one term at a time.
what is:\[_nP_6=?\]in terms of factorials?
OpenStudy (thatonegirl_):
n!/(n-6)!
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OpenStudy (asnaseer):
good, now what is:\[_nP_5=?\]in terms of factorials?
OpenStudy (thatonegirl_):
n!/(n-5)!
OpenStudy (asnaseer):
great. your equation is therefore:\[_nP_6=5\left(_nP_5\right)\]\[\therefore \frac{n!}{(n-6)!}=5\left(\frac{n!}{(n-5)!}\right)=\frac{5\times n!}{(n-5)!}\]
OpenStudy (asnaseer):
the \(n!\) terms on both sides can be cancelled out
OpenStudy (thatonegirl_):
ohh ok gotcha
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OpenStudy (asnaseer):
then use the hint I gave you earlier and you should be able to solve this fairly quickly :)
OpenStudy (thatonegirl_):
isnt it still 5/(n-6)(n-5) ?
OpenStudy (asnaseer):
how did you get that?
OpenStudy (thatonegirl_):
multiplying both equations
OpenStudy (asnaseer):
please use the drawing tool to show your steps if you cannot type in latex
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OpenStudy (thatonegirl_):
\[n!(n-5)!=(n-6)!(5\times n!) \]
OpenStudy (thatonegirl_):
ya idk
OpenStudy (asnaseer):
correct so far :)
now you can cancel the \(n!\) from both sides
OpenStudy (asnaseer):
leaving you with:\[(n-5)!=5(n-6)!\]
OpenStudy (asnaseer):
then use the fact that:\[(n-5)!=(n-5)(n-6)!\]and you get:\[(n-5)(n-6)!=5(n-6)!\]now you can cancel out the \((n-6)!\) terms from both sides
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