We can arrange all the 7 non-blue balls in 7! ways.

Below, think of the 7 X's as representing the 7 non-blue balls.
Think of the 8 blanks as representing the 8 positions any 5 in 
which the 5 blue balls could be placed, between two non-blue
balls, left of the first non-blue ball or right of the last
non-blue ball.

_X_X_X_X_X_X_X_

For each of the 7! ways the non-blue balls could be placed,
there are P(8,5) ways to place the 5 blue balls.

That's (7∙6∙5∙4∙3∙2∙1)(8∙7∙6∙5∙4) = ways they could be placed
out of 12! ways the balls could be placed with no restrictions:


(7∙6∙5∙4∙3∙2∙1)(7∙6∙5∙4∙3)                     7
--------------------------- = after canceling ---
12∙11∙10∙9∙8∙7∙6∙5∙4∙3∙2∙1                     99

Edwin