Ih how many ways 21 red balls and 19 blue balls can be arranged in a row so that no two blue balls are together

I'll give you a hint to get you started, but will leave it to you to finish it up.

The 19 blue balls must each have at least 1 red ball between them, so that uses up 18 of the reds. Which means there are three left over, each one of which can go with any one of the first 18, or at the beginning of the line (before the first blue ball), or at the end (after the last blue ball). That's 20 total slots into which these last these 3 balls can be placed.

Now, can you take it from here?