I have the following set: $$\{1,2,3,4,5,6,7,8,9,10\}$$
I want to split this set into two equal groups (e.g group of 5) and I have been asked how many ways are there to split this set when the order doesn't matter (e.g if group $\{1,2,3,4,5\}$ is the same case as $\{5,4,3,2,1\}$.
I think the beginning of the answer is to $2^{10}$ for the number of options but then in order to reduce the cases where only the spot of each elemnt is different I got confused and not sure that this is the way.
I'm very beginner with this so might be that also the $2^{10}$ is wrong.
So my wurstion is how can I calculate the number of ways to split this set into 2 groups without taking into accound the order of elemnts?
Edit: it does not matter which of the subsets is the first and which is the second
Best Answer
The question asks you to select 5 random numbers in one group and the remaining 5 in the other group. For selecting the first 5 elements, we use combinatorics:
no. of ways of selecting first 5 numbers = 10C5 = 252
and for the remaining 5 numbers no. of ways = 5C5 = 1
final answer = (252*1)/2 = 126
(this is because : {1,2,3,4,5},{6,7,8,9,10} is the same as {6,7,8,9,10},{1,2,3,4,5})