SOLUTION: An urn contains 4 red, 2 white, and 2 blue marbles. In how many ways can three marbles be selected so that at least one of them is blue?

There are two methods:

First method:

We add the number of ways defined by these cases:

Case 1:  We pick both blue balls and one non-blue ball

Case 2:  We pick 1 blue ball and 2 non-blue balls

For case 1, we pick both blue balls 1 way, then the other ball,
any of 6 ways.  That's 1x6 = 6 ways

For case 2, we pick the blue ball either of 2 ways, then the 
2 non-blue ball, any of 6C2 ways.  That's 2x15 = 30 ways.

So the total number of wasy is 6 + 30 = 36

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Second method:

The number of ways of picking ANY 3 balls from the 8 balls

             MINUS

the number of ways of picking 3 NON-blue balls from the 6 NON-blue balls.

That's 8C3 - 6C3 =  56 - 20 = 36 ways.

Either way gives the same answer.

Edwin