$24, 2$ red balls and $11$ white balls are arranged in a row, satisfying the following conditions: the red balls are not adjacent, and in any consecutive $7$ balls, there is at least one red ball, then there are $\qquad$ ways to arrange them.
【Analysis】If there is 1 white ball between two red balls, the remaining 10 white balls can be placed on the left and right in (4,6) (5,5) (6,4), a total of 3 arrangements.
If there are 2 white balls between two red balls, the remaining 9 white balls can be placed on the left and right in (3,6) (4,5) (5,4) (6,3), a total of 4 arrangements.
If there are 3 white balls between two red balls, the remaining 8 white balls can be placed on the left and right in (2,6) (3,5) (4,4) (5,3) (6,2), a total of 5 arrangements.
If there are 4 white balls between two red balls, the remaining 7 white balls can be placed on the left and right in (1,6) (2,5) (3,4) (4,3) (5,2) (6,1), a total of 6 arrangements.
If there are 5 white balls between two red balls, the remaining 6 white balls can be placed on the left and right in (0,6) (1,5) (2,4) (3,3) (4,2) (5,1) (6,0), a total of 7 arrangements.
If there are 6 white balls between two red balls, the remaining 5 white balls can be placed on the left and right in (0,5) (1,4) (2,3) (3,2) (4,1) (5,0), a total of 6 arrangements.
In total, there are 31 arrangements.