nCr question choosing 1 - 9 from 9 I've been trying to rack my brain for my high school maths to find the right calculation for this but I've come up blank.
I would like to know how many combinations there are of choosing 1-9 items from a set of 9 items.
i.e.
There are 9 ways of selecting 1 item.
There is 1 way of selecting 9 items.
For 2 items you can choose...
1,2
1,3
1,4
1,5
1,6
1,7
1,8
1,9
2,3
2,4
and so on.
How many ways in total are there of selecting any number of items from a set of 9 (without duplicates, i.e. you can't have 2,2).
Also, order is not important. So 1,2 is the same as 2,1.

Line up the items in front of you, in order. To any of them you can say YES or NO. There are $2^9$ ways to do this. This is the same as the number of bit strings of length $9$.
But you didn't want to allow the all NO's possibility (the empty subset). Thus  there are $2^9-1$ ways to choose $1$ to $9$ of the objects.
Remark: There are $\dbinom{9}{k}$ ways of choosing exactly $k$ objects. Here $\dbinom{n}{k}$ is a Binomial Coefficient, and is equal to $\dfrac{n!}{k!(n-k)!}$. This binomial coefficient is called by various other names, such as $C(n,k)$, or ${}_nC_k$, or $C^n_k$.
So an alternate, and much longer way of doing the count is to find the sum
$$\binom{9}{1}+\binom{9}{2}+\binom{9}{3}+\cdots +\binom{9}{9}.$$
