number of combination in which no two red balls are adjacent.

Congratulations on the experimentation! That is always a good beginning. The question is an old standard, indeed, because of the connection with Fibonacci numbers, a golden oldie. Call an arrangement of balls good if there are never two red balls in a row.

Let $f(n)$ be the number of good arrangements of $n$ balls. How many good arrangements of $n$ balls end in a white? Clearly there is exactly one such good arrangement for every good arrangement of $n-1$ balls. For from every good arrangement of $n-1$ balls, we can get a good arrangement of $n$ balls ending in white by appending a white. And from every good arrangement of $n$ balls ending in white, we can get a good arrangement of $n-1$ balls by removing the last white. So the number of good arrangements of $n$ balls that end in a white is $f(n-1)$.

How many good arrangements of $n$ balls end in a red? The previous ball must be white. And by the preceding paragraph, there are just as many good arrangements of $n-1$ balls that end in white as there are good arrangements of $n-2$ balls (of course, we need $n-2\ge 0$). Since every good arrangement of $n$ balls either ends in a white or a red, we conclude that $f$ satisfies the recurrence $$f(n)=f(n-1)+f(n-2).$$ The initial conditions are $f(0)=1$ and $f(1)=2$. We get essentially the Fibonacci sequence, with slightly modified indexing. If (as in the Wikipedia article linked to) we use $F_0=0$, $F_1=1$ for the Fibonacci sequence, then $f(n)=F_{n+2}$.