I need to calculate the events corresponding to the number of ones in 10 dice rolls.
When $X$ is a r.v. for the number of ones, then this produces events:
$(X=0)=(0,0,...,0)$
$(X=1) = \{(1,0,...,0), ..., (0,0,...,1)\}$
...
$(X=10)=\{(1,1,...,1)\}$
Listing all these as sets proves to be slightly challenging.
So is there a formula for finding the permutations in these sets or specifically the number of permutations in each set (in order to be able to formulate the probability distribution of $X$)?
For $X=0$ and $X=10$ the answer is clear. In the other cases we are talking about permutations with repetition. The formula for $X=k$ is $$\dfrac{10!}{k!(10-k)!}.$$
Note that if $0'$s and $1'$ are labelled then the number of possible positions is $10!.$ But since we do not distinguish between $1'$ we must divide by $k!,$ wich is the number of permutations of $1'$s if they were labelled. For the same reason applied to $0'$s we divide by $(10-k)!.$
We have to note that $$\dfrac{10!}{k!(10-k)!}=\binom{10}{k}.$$ (It works also for $k=0,10.$) We have a formula form the sum. It is
$$2^{10}=(1+1)^{10}=\sum_{k=0}^{10}\binom{10}{k}.$$