$n$ balls numbered from $1$ to $n$ 
If we consider a box containing $n$ balls numbered from $1$ to $n$. If $455$
  is the number of ways to get three balls from the box such that no two
  balls are consecutively numbered, then we have to find the value of $n$.

Someone please help me out in this. I am not getting anything, how to start it?

The general problem at hand is this: How many ways are there to choose $k$ out of $n$ numbered balls such that no two chosen balls have consecutive numbers?  
To approach this, think about lining up the $n-k$ balls that are not chosen.  Then insert the chosen balls between the unchosen balls, where they belong in the line.  If no two chosen balls are consecutive, each chosen ball gets stuck between a different pair of unchosen balls.  Conversely, any way of sticking $k$ chosen balls between $n-k$ unchosen balls, such that no two chosen balls are between the same pair of unchosen balls, gives a choice of $k$ non-consecutive balls from the total $n$ balls.  
So we see that the number of ways to choose $k$ non-consecutive balls out of $n$ total balls is precisely the number of ways to put $k$ chosen balls between $n-k$ unchosen balls, with at most one chosen ball between any two unchosen ones.  Note that we can also put a chosen ball on the far left or far right of the line of unchosen balls, so the total number of places to put the chosen balls is $n-k+1$ (explicitly, the chosen balls can be to the left of the first unchosen ball, to the left of the second, to the left of the third, ..., to the left of the last unchosen ball, or to the RIGHT of the last unchosen ball). 
Thus the number of interest is the number of ways to put $k$ chosen balls in $n-k+1$ locations, i.e. $\binom{n-k+1}{k}$.  
For the problem of interest, $k=3$, and we want to know for which $n$ is $455=\binom{n-3+1}{3} = (n-2)(n-3)(n-4)/6$.  This is a polynomial you can solve (or, more likely, just check a few cases), and it has solution $n=17$. 
