I want to count the number of ways to arrange the digits $1,2,3,4$ and $5$ such that $1$ is to the left of $2$. I can think of this as choosing two positions for $1$ and $2$, and then filling the remaining three positions with $3,4$ and $5$ in any order. How many ways can I choose two positions for $1$ and $2$ out of five? I can use the combination formula: $\binom{5}{2} = \frac{5!}{2!3!} = \frac{5\times 4}{2} = 10$. So there are $10$ ways to choose where $1$ and $2$ go, and for each of these choices, there are $3! = 6$ ways to arrange the remaining digits. Therefore, the total number of ways to arrange the digits $1,2,3,4$ and $5$ such that $1$ is to the left of $2$ is $10 \times 6 = 60$. 
The digits $1,2,3,4$ and $5$ can be arranged to form many different $5$-digit positive integers with five distinct digits. In how many such integers is the digit $1$ to the left of the digit $2$?  (The digits 1 and 2 do not have to be next to each other.)