Bunuel wrote:

Official Solution:

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

Total ways to select 8 marbles out of \(7+5=12\) is \(C^8_{12}\);

Ways to select 8 marbles so that zero red marbles is left in the jar is \(C^7_7*C^1_5\);

Ways to select 8 marbles so that zero blue marbles is left in the jar is \(C^5_5*C^3_7\);

Hence ways to select 8 marbles so that at least one red marble and at least one blue marble remain in the jar is \(C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455\).

Answer: D

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!