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Mathematics 25 Online
OpenStudy (thatonegirl_):

Solve for n

OpenStudy (thatonegirl_):

\[_nP_6=5(_nP_5)\]

OpenStudy (anonymous):

is this algebra 2?

OpenStudy (asnaseer):

First write these in terms of factorials

OpenStudy (asnaseer):

remember that:\[_nP_r=\frac{n!}{(n-r)!}\]

OpenStudy (thatonegirl_):

ya @osamabinswaggin

OpenStudy (thatonegirl_):

i got tht @asnaseer

OpenStudy (asnaseer):

so where exactly are you stuck on this?

OpenStudy (asnaseer):

HINT: \((n-5)!=(n-5)(n-6)(n-7)...(3)(2)(1)=(n-5)(n-6)!\)

OpenStudy (thatonegirl_):

do i just plug in what i know?

OpenStudy (thatonegirl_):

and where does the 5 outside of the (nP_5) come from??

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)!

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

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

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

OpenStudy (thatonegirl_):

where did u get the (n-5) on the right side?

OpenStudy (asnaseer):

\((n-5)!=(n-5)(n-6)(n-7)...(3)(2)(1)=(n-5)[(n-6)(n-7)...(3)(2)(1)]\) \(=(n-5)[(n-6)!]\)

OpenStudy (thatonegirl_):

oh ok

OpenStudy (asnaseer):

e.g.\[8!=8*7*6*5*4*3*2*1=8*[7*6*5*4*3*2*1]=8*7!\]

OpenStudy (thatonegirl_):

ohh yea I get it now

OpenStudy (asnaseer):

good - so hopefully you can now solve the equation you were given?

OpenStudy (thatonegirl_):

n=10?

OpenStudy (asnaseer):

perfect! :)

OpenStudy (thatonegirl_):

thank you so much!

OpenStudy (asnaseer):

yw :)

OpenStudy (anonymous):

i completely forgot about this...I went off doing other things and lost the medal *cries*

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