Question 1018796: In a box there are: 7 red books, 5 white books, and 6 blue books. Three books are selected.
How many different ways can this be done if the selection:
(a) must contain one of each color? c(7,1)c(5,1)c(6,1)= 210
(b) must contain all the same color? c(7,3)+c(5,3)+c(6,3)=65
(c) must include more than one color? I'm not sure how to set this up can you please guide me through?
Found 2 solutions by Edwin McCravy, robertb:
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website! (c) must include more than one color?
There are 7+5+6 = 18 books.
We can choose 3 book any of c(18,3) = 816 ways.
But, from those 816 ways to choose any 3 books, we
must subtract the number of ways they were all
the same color, which you found correctly to be
65 in part (b). c(7,3)+c(5,3)+c(6,3)=65
Answer: 816-65 = 751 ways.
Edwin
Answer by robertb(5830)  (Show Source):
You can put this solution on YOUR website! C(18,3) = 816 is the number of ways of selecting any three books from the eighteen books without any restriction.
Now the number of ways in which one book of each color is selected is 210, which you have correctly determined. The complement of this event is the event in which AT LEAST one book of some color is selected.
Thus, the number of ways in which a selection includes more than one color is
816 - 210 = 606.
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