Probability with sets The numbers from 1 to 10 are divvied randomly into 2 sets of 5 elements each. 
What is the probability that numbers 1 and 2 are in the same set?
My attempt is below:
The total number of ways to divvy 10 numbers into two sets of 5 elements is $\frac{10!}{5!5!}$
The number of ways in which numbers 1 and 2 are in the same set is = $\dbinom{8}{3}+\dbinom{8}{3}$
 A: In Species language, the structure in the problem is 
 a 2-set of 5-sets of A,B, and X 
where A and B are constants.
By definition, C is a constant $ \iff F(C,X) = C \cdot F'(X) $
Here we have:
$ E_2(E_5(A, B, X)) = A \cdot B \cdot E_2(E_5(X))'' $
$ = A \cdot B \cdot E_5E_4(X)' = A \cdot B \cdot  (E_5E_3 + E_4E_4)(X)'$
$ = E_5(X) \cdot E_5(A,B,X) + E_4(A,X) \cdot E_4(B,X) $
The e.g.f. of above is 
$$ \left [ {8 \choose 5,3} + {8 \choose 4,4} \right ] ab { x^8 \over 8!} $$
thus, there are 56 good configurations and 70 bad ones.
Then we have to apply the definition of a probability.
A: What you did is correct. There are $ {10 \choose 5}$ ways to divvy up the numbers 1 to 10 into 2 sets. The number of ways for 1 and 2 can be in the first set is ${8 \choose 3}$, since we can place 1 and 2 in the first set and then fill in another 3 elements from the 8 remaining ones. Similarly, there is also ${8 \choose 3}$ ways of putting 1 and 2 into the second set, and therefore $2 {8 \choose 3}$ ways to have 1 and 2 in the same set.
Hence, the probability of having 1 and 2 in the same set in this scenario is: 
$$
\dfrac{2 {8 \choose 3}}{{10\choose5}} = \dfrac{4}{9}
$$ 
A: That is okay, but consider a simpler way.
The numbers are divided evenly into sets of size five.  That is ten places for the numbers to occupy.
Number 1 will be in one of these places and four of the nine remaining places will be in the same set.
The probability that number 2 in placed among those four of the nine places is ………… ?
