Combinatorics question regarding arranging $13$ balls in a line but with a condition that on neither of the sides have a blue ball. I have $13$ balls. $6$ red , $4$ blue , $3$ yellow. I want to arrange in a line such that the right side and the left side do not have blue balls. ( balls in the same color are not distinct ).
I tried to calculate the options without any terms or conditions which is 13! divided by $6!4!3!$. ( ! is factorial )
and then subtract the options that contain having blue ball on the right side and the left side which is : ($4$ choose $2$ for the blue balls on each side) multiply by ( $11!$ divided by $(6!2!3!)$ which are the remaining $11$ balls left).
but I'm not getting the right answer. What am I doing wrong in the process?

Blue balls are identical so 4c2 is not needed . The cases should be in the order
(i) Only blue on left and not on right
(total case where blue is on left - case where blue on both left and right)
(ii) Only blue on right and not on left
(total case where blue is on right - case where blue on both left and right)
(iii) Blue on both left and right .
Try calculating it
Hope this helped
