There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble remain in the jar?

A. 460
B. 490
C. 493
D. 455
E. 445
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Bunuel,

This is a duplicate question. Please refer to M13-37.

The question says at least one red marble AND at least one blue marble.
But you add the possibilities (5+35). Shouldn't it be multiplied ?

Total ways of selecting 8 marbles=495
Marbles that remain in the jar= 1R and 3B or 3R and 1B= 35/2 *2= 35. In this case shouldn't the answer be 460?
Am I missing something?

Bunuel
Are the same colored marbles identical?
I didn't understand how there would be 5 ways to choose a blue marble if they are all identical? I thought there should be 1 way only if the red marbles are identical to each other and blue identical to the other blue.

thank you

The question prompt states:"at least one red marble AND at least one blue marble." Yet, the answer seems to consider one red marble OR one blue marble.

Because when you take out the possibility of there being either zero Reds or zero Blues from all available options, it gives you all the other ways (in numbers) where after taking out 8 marbles, there is at least 1 Red and at least 1 Blue left in the jar.
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Just to explain this a bit more:

Problem says there are 12 marbles total, 7 red 5 blue.
Problem wants to select 8, so that there are at least 1 red and 1 blue not selected.

When you see "at least" (or "at most") this is a sign that you want to consider doing P(A) - P(not A) to get your answer.
So consider how many ways there are to select 8 marbles out of 12, with no restrictions. This is 12C8, or 12!/4!8! = 495.

Then think about what the problem wants. What is the opposite of the situation in which you have at least one red and at least one blue? The opposite is you have no red and no blue. That is the opposite of at least 1 red and at least 1 blue.

With that in mind, think about how you would get into those situations... and note that these are two separate (OR) situations.
To have zero red left, you need to pick 7 red and 1 blue.
To have zero blue left, you need to pick 5 blue and 3 red.

Convert these both to math...
Zero red = 7C7*5C1 = 7!/7! * 5!/1!4! = 5
Zero blue = 5C5 * 7C3 = 5!/5! * 7!/3!4! =35
Add the two together and you get 40.

Now remember P(A)-P(not A) = answer.
Do the math:
495 - 40 = 455

455 is the answer!

Hi Bunuel
Could you please tell me why I got a wrong answer ?
First ,we select one red , one blue , we left them in the jar : number of ways of doing this : 5*7=35
this way we ensure that this is fixed , one red and one blue are in the jar .
then , 10 are left , we choose 8 from them , number of ways : C(10,8) : 45
The total number of ways : 35*45 = 1575
and we are sure that at least one red , one blue are in the jar .

What is wrong ?
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