A formula for finding the number of strings obtained by arranging \(K\) K’s, \((R-K)\) R’s, \((G-K)\) G’s, and \(B\) B’s, that does not contain a substring RG can be derieved.

First arrange R, RG and B. There are \(\frac{(b + k + r-k)!}{(r-k)!b!k!}\) ways of doing so. Now all the G’s must be filled in the gaps. We can’t fill in G afterR because then they will make RG. So the number of spots where we can fill in G’s are \(((r-k) + b + k + 1 - (r-k)) = b + k + 1\). So the final answer can be given by the formula,

\(\frac{(b + k + r-k)!}{(r-k)!b!k!} \cdot \binom{(g-k) + spots - 1}{g-k}\)

Where \(spots = b + k + 1\).