Combinatorics problem! (Probably easy)

Let $n_{1}, n_{2}, n_{3}$ be a given combination of $3$ integers chosen, and without loss of generality suppose $n_{1} \lt n_{2} \lt n_{3}.$

Now let $a$ be the number of integers in $S = {1,2,3,....,49}$ less than $n_{1},$ $b$ the number of integers in $S$ between $n_{1}$ and $n_{2},$ $c$ the number of integers in $S$ between $n_{2}$ and $n_{3},$ and $d$ the number of integers in $S$ greater than $n_{3}.$

Then $a + b + c + d = 46$ such that $a,d \ge 0$ and $b,c \ge 1,$ (in order to ensure that $n_{1}, n_{2}, n_{3}$ are not consecutive).

Now let $b' = b - 1$ and $c' = c - 1.$ We then have the equation $a + b' + c' + d = 44$ with $a,b,c,d \ge 0,$ which is a 'stars and bars' calculation with solution $\dbinom{47}{3} = \boxed{16215}.$

In similar fashion, the general formula is found to be $\dbinom{n - r + 1}{r}.$

let us have n numbers and we need to choose r out of them so let us have r markers for the numbers we choose and n-r markers for the numbers we don't choose. so those n-r markers are arranged in one way and they have n-r+1 spaces in between them for us to insert those markers of numbers we want to choose so total ways in which we can choose r spaces out of n-r+1 spaces = $\dbinom{n-r+1}{r}.$ after the combination has happened i will lift up the whole pattern and lay it on the line of numbers...