SOLUTION: A bowl contains 9 red balls and 9 blue balls. A woman's selects 4 balls at random from the bowl. How many different selections are possible if at least 3 balls must be blue ?
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Question 988923: A bowl contains 9 red balls and 9 blue balls. A woman's selects 4 balls at random from the bowl. How many different selections are possible if at least 3 balls must be blue ? Answer by Shai(25) (Show Source):
You can put this solution on YOUR website! Given:
Total number of red balls : 9
Total number of blue balls : 9
Also a woman selects 4 balls
And also at least 3 balls off the 4 must
Be blue
So first combination is 3 blue and 1 red
(remember the woman picks up 4 balls)
And the second combination is 4 blue and 0
Red(remember at least 3 blue balls)
So the equation is as given below:
(9C3*9C1 + 9C4*9C0)=(84+126)
Your answer is 210
