Picking $6$ numbers from $\{1, \ldots, 49\}$, what is the chance that the difference between at least $2$ of them is $= 1$? You choose $6$ different natural numbers from $\{1, \ldots, 49\}$. What is the probability that at least $2$ of these numbers have a difference equal to $1$? 
E.g.   $1, 2, 10, 20, 30, 31$ - you'd have $2$ pairs $(1,2), (30,31)$ with the difference  $= 1$. 
I tried solving it by taking the inverse probability $P(\neg A)$ < No $2$ numbers have a difference of $1$ >, however, I encounter some difficulties with some special cases. For the first number, you'd have $49$ options. For the second number however, you'd have $46$ options (removing the previous number, and the $2$ surrounding it - which would give a diff. of $1$). However if the first number was $1$ or $49$, you'd have $47$ options for the your second number, as there could only be $1$ number that would give you a diff of $1$ ($2/48$). It only gets more complicated if two numbers are $x$ and $x+2$, because the next number cannot be $\{x-1,x+1,x+3\}$.
Thank you.
 A: To count the number of arrangements in which no two selected numbers are consecutive, we will arrange $43$ blue and $6$ green balls in a row so that no two of the green balls are adjacent.  Line up the $43$ blue balls.  This creates $44$ spaces in which we could place a green balls, $42$ between successive blue balls and two at the ends of the row.  To ensure that no two green balls are consecutive, we must choose six of these $44$ spaces in which to insert a green ball.  Once that choice is made, number the balls from left to right.  The numbers on the green balls are the desired set of numbers of which no two are consecutive.  Since there are $\binom{49}{6}$ ways to select six of the $49$ numbers, the probability that no two of the selected numbers are consecutive is 
$$\Pr(\text{no two are consecutive}) = \frac{\dbinom{44}{6}}{\dbinom{49}{6}}$$
Therefore, the probability that at least two of the selected numbers are consecutive is
$$\Pr(\text{at least two are consecutive}) = 1 - \frac{\dbinom{44}{6}}{\dbinom{49}{6}}$$
