How many ways are there to put 10 red and 9 blue balls in a sequence so that for every index the number of red balls up to and including this ball is greater than the number of blue balls?
It means that the first two balls must be red, because two red balls are required to place one blue ball, and the last ball must be blue, because all 10 red balls have to be placed before the 9th blue ball can be placed.
We need the number of red balls strictly greater than the number of blues balls at any given index. Like you said this means the first ball must be red. Place the first red ball then forget about it. After this we have 18 balls to place and 9 balls of each color, and with in just these 18 balls the number of red balls in a given place must be greater than or equal to the number of blue balls. The number of ways to do this is $C_9 = \frac{1}{10}\binom{18}{9}$ where here $C_9$ denotes the 9th Catalan Number. In the Catalan Number link check the "Applications in combinatorics" sections if your are note familiar with this fact. It is not hard to come up with a bijection between what you are counting and some of the objects listed in this section (for example Dyck Words).