Question 477438: 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 stanbon(75887)   (Show Source):
You can put this solution on YOUR website! "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,
Easiest Way:
# of ways to get 3 red: 7C3 = (7*6*5)/(1*2*3) = 35
------------------------
b) none are red?
Easiest Way:
# of ways to get 3 not red: 10C3 = (10*9*8)/(1*2*3) = 120
--
Your Question::::::
17C3 - 7C3 = 680-35 = 645
The 645 includes all the combinations that allow
1 red in the group of 3 or 2 reds in the group of 3.
---
Your 35 is the # of groups with ALL red
Your 120 is the # of groups with NO red
============================================
Cheers,
Stan H.
|
|
|
