Question

MEDAL MEDAL MEDAL.
Factorial with variables

Answer 1

?

Answer 2

\[\left(\begin{matrix}n+1 \\ n-1\end{matrix}\right)\]

Answer 3

what are you trying to do?

Answer 4

if you write out the first few terms it should be obvious more or less that
\[\frac{(n+1)!}{(n-1)!}=(n+1)n\]

Answer 5

so it'll be \[\frac{ (n+1)! }{ [(n+1)-(n-1)!](n-1)! }\]

Answer 6

wait what are you talking about

Answer 7

yeah i see
pretty sure it is just
\[\frac{n(n+1)}{2}\]

Answer 8

lets do some algebra first ok?

Answer 9

yeah that was correct. thanks so much. the answer was \[\frac{ n(n+1) }{ 2 }\]

Answer 10

lol yeah i know
do you see how i got it?

Answer 11

not really

Answer 12

just takes a little practice working with factorials and you see how things cancel

Answer 13

first of
\[(n+1)-(n-1)=n+1-n+1=2\] in the denominator and \(2!=2\)

Answer 14

as for
\[\frac{(n1)!}{(n-1)!}\] if you write out the first few terms of the numerator you get
\[(n+1)n(n-1)(n-2)(n-3)...=(n+1)n(n-1)!\]

Answer 15

the \((n-1)!\) cancels with the denominator and you are just left with
\[\frac{n(n+1)}{2}\]

Answer 16

like for example
\[\frac{7!}{5!}=7\times 6\]

Answer 17

oh okay. thanks so much @satellite73