Question 1165592: Given three red balls, four green balls and five blue balls, what is the probability that when you arrange those balls, no blue balls
are adjacent to each other?
a. 1/99 b. 4/99 c. 5/99 d. 7/99
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website! 3 red balls, 4 green balls and 5 blue balls,
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
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