The combinations of four distinct digits that sum to 10 are $0127, 0136, 0145, 0235, 1234$, so the total number of four-digit numbers is $18 \cdot 4 + 24 = 96$. Let $A=$ “2017 is selected”, so $P(A)=\frac{95}{C_{96}^{2}}=\frac{1}{48}$.
7. From all four-digit numbers where each digit is unique and the sum of the digits is 10, if two numbers are randomly selected, what is the probability that 2017 is selected?