Question 477545: Dear math teacher,
I am having the following difficulties with this problem:
"A box contains 7 red cards, 6 white cards and 4 blue cards. How many selections of three cards can be made so that a) all three are red, b) none are red?"
I solved the problem correctly but then I tried to solve it the second way - the shorter and less time consuming way. I was supposed to get the same result but did not. Please let me know, why I did not get the same answer.
Here is what I did by Method 1 for part a):
R W B
n 7 6 4
r 3
7C3 = 35 selections
Here is what I did by Method 1 for part b):
R W B
n 7 6 4
r n = 6+4 = 10
r = 3
10C3 = 10!/(10-3)! = 120 selections.
Method 2 part b)
n (total R, W, B) = 7+6+4 = 17
r = 3
17C3 - 7C3 = 680-35 = 645 selections and this answer does not match the answer obtained by Method 1 for part b). Would you tell me why? Where did I do a mistep? Here is how reasoned it out: Total Number of Ways - Number of Ways to Pull Red Cards = Number of Ways to Get Other Colors
Why would that be wrong to think like that?
Thank you very much for your help.
Respectfully,
Ivanka
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! the two groups are "all red cards" and "no red cards"
in method 2 you found all possible ways to select 3 cards; and then subtracted all the ways to get 3 red cards
___ this does not account for groups containing one or two red cards
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