Question

Find all possible values of non negative integers n such that:

Answer 1

\[\left(\begin{matrix}n+2 \\ 5\end{matrix}\right)=\frac{21}{5}\left(\begin{matrix}n \\ 4\end{matrix}\right)\]If this is the question then I can't find ANY integer solutions to it.
Are you sure the question is correct?

Answer 2

yes...

Answer 3

how do i approach such a question though?

Answer 4

\[\begin{align}
\left(\begin{matrix}n+2 \\ 5\end{matrix}\right)&=\frac{21}{5}\left(\begin{matrix}n \\ 4\end{matrix}\right)\\
\frac{(n+2)!}{5!(n+2-5)!}&=\frac{21}{5}\times\frac{n!}{4!(n-4)!}
\end{align}\]
if you continue from here and simplify you end up with:\[(n+2)(n+1)(n-4)=21\]which has no integer solutions. unless I made a mistake somewhere?

Answer 5

I'm getting two solutions.

Answer 6

I'm taking:\[\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{n!}{r!(n-r)!}\]

Answer 7

hmmm. ok...could it be a trick question?

Answer 8

Start with \[\frac{(n+2)!}{5!(n-3)!}\overset{?}=\frac{21}{5}\times\frac{n!}{4!(n-4)!}\]Simplify the left side and get a common denominator to get \[\frac{21(n-3)n!}{5!(n-3)!}\]Now you want to find integers \(n\) such that \[(n+2)!=(21n-63)n!\]Divide out by \(n!\) to get \((n+1)(n+2)=21n-63\).

Answer 9

If I've done everything correctly, that should have two integer roots.

Answer 10

are you sure that is correct KingGeorge?

Answer 11

Pretty sure since it kicks out two integer values as roots, and the roots seem to work with the original question.

Answer 12

I got the same thing as KingGeorge here.

Answer 13

ah! I see where I made the mistake, I said:\[(n-4)!=(n-4)(n-3)!\]which is clearly wrong!

Answer 14

Those pesky minus signs screw everything up.

Answer 15

indeed :)

Answer 16

im stuck with \[\left(\begin{matrix}n+2 \\ 5\end{matrix}\right)\]

Answer 17

?

Answer 18

@Keroro do you agree that:\[\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{n!}{r!(n-r)!}\]

Answer 19

yes

Answer 20

then you must agree that:\[\left(\begin{matrix}n+2 \\ 5\end{matrix}\right)=\frac{(n+2)!}{5!(n+2-5)!}=\frac{(n+2)!}{5!(n-3)!}\]

Answer 21

yes and then i dont knw how to get further than that..

Answer 22

\[\begin{align}
\left(\begin{matrix}n+2 \\ 5\end{matrix}\right)&=\frac{21}{5}\left(\begin{matrix}n \\ 4\end{matrix}\right)\\
\frac{(n+2)!}{5!(n+2-5)!}&=\frac{21}{5}\times\frac{n!}{4!(n-4)!}\\
\frac{(n+2)!}{5!(n-3)!}&=\frac{21n!}{5*4!(n-4)!}=\frac{21n!}{5!(n-4)!}=\frac{21n!}{5!(n-4)!}\times\frac{n-3}{n-3}=\frac{21(n-3)n!}{5!(n-3)!}\\
\frac{(n+2)(n+1)\cancel{n!}}{\cancel{5!(n-3)!}}&=\frac{21(n-3)\cancel{n!}}{\cancel{5!(n-3)!}}\\
(n+2)(n+1)&=21(n-3)\\
\end{align}\]

Answer 23

does that make it clearer?

Answer 24

oh yes, i see, i had to times the n-3/n-3 thing

Answer 25

Thanks so much everybody:)

Answer 26

yes as @KingGeorge showed above :)

Answer 27

yw :)

Answer 28

I glossed over that step a bit in my explanation :/