By replacing RG with K in a string satisfying the conditions, the problem is rephrased as:

Find 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.

We introduce two approaches to the rephrased problem.

[1] Inclusion-exclusion principle

It can be solved by applying inclusion-exclusion principle with respect to the number of RG’s. The number of strings containing at least \(i\) RG’s is obtained as a multinomial coefficient. The complexity is \(\mathrm{O}(R+G+B)\).

Sample code (c++)

[2] Combinatorial interpretation

First, freely arrange K,G, and B. Then we can insert R to any position except for inserting in the left of G. Both can be found easily as binomial and multinomial coefficients. The complexity is \(\mathrm{O}(R+G+B)\).

By the way, with a fast algorithm to find a factorial, we can solve it in a total of \(\mathrm{O}(\sqrt P\log P)\) time, where \(P=998244353\).

Sample code in \(\mathrm{O}(R+G+B)\) (c++)