Question

I am trying to figure out an appropriate method in determining how many terms there are for any given itteration

So basically We are given an example where any particular stock can either go up by a factor or down by a factor after every closing period. So for instance after 5 closing periods how many term will there be taking into consideration that order obv doesn't matter since UUD=DUU=UDU

Answer 1

I believe the number you're looking for is perm(x)/x!
Does that sound familiar?

Answer 2

what does perm(x) mean?

Answer 3

ohhh permutation

Answer 4

Or wait, \[\frac{\left(\begin{matrix}n \\ 2\end{matrix}\right)}{n!}\]

Answer 5

Right? I think dividing by n! covers the issue with order.

Answer 6

Haha whatever I'm doing is way off, but conceptually close I think. Obviously my number is less than 1.

Answer 7

ya no that wont work
lemme draw a graph for the 3rd iteration

Answer 8

S represents the stock
u=up factor
d=down factor

Answer 9

So for the third iteration you have 4 choices

Answer 10

Could it be \[\frac{2^n}{(n-1)!}\]

Answer 11

hmmmmmmm

Answer 12

No, that's still not right for the 2nd iteration. It's right for 1 and 3 though. Weird.

Answer 13

kai you lil bish ...

Answer 14

So it's definitely going to be 2^n minus something. What you have is n spots, each with two options. Hence 2^n. But because of repetition, you need a way to count those that are repeated.

Answer 15

lol I'm not totally sure what you're asking for but like there will always be only n+1 possible choices and they will be distributed like @escolas is saying.

Answer 16

Haha wow, right..

Answer 17

omg -.-

Answer 18

this is embarrassing

Answer 19

Hahahaha this is called "Pascal's Triangle" lol

Answer 20

hahaha

Answer 21

This isnt even funny ,,, this is just sad

Answer 22

Can I erase all my embarrassing answers?

Answer 23

thanks escolas for attempting to help :)

Answer 24

ya sure :)

Answer 25

Hahah any time.. :P

Answer 26

Haha not sure I deserved that medal, but thank you very much. Cheers.