Probability Question

[vsudheerk]

How would you answer the following question:

How many different combinations of 3 balls can be picked from a bag containing 10 Red, 10 Green, and 10 Blue balls? (Note that RGB = RBG = BRG ...., RRB=BRB=BBR, etc..)

Thanks
SV

[JonathanSchneider]

As we are not interested in the number of possible orderings, this problem has been greatly simplified. We only need to determine the number of ways to choose 3 items.

There are only three ways to choose in general:

All three colors the same
Two of the same color, one of a different color
All three colors different

Now, let us count the number of options for each:

All three the same: 3 ways to do this, one for each of the three colors

Two of the same color, one of another: We are excluding one color; there are three ways to choose which color to exclude. Once we have chosen the other two colors, we must choose which repeats, and we have two choices. Thus, there are a total of 6 ways to create a two-and-one set.

All three different: only one way to do this, as we need one of each.

Thus: 3 + 7 + 1 = 11 total ways.

[jainshweta4]

Well I think the answer should be 10 not 11 as per ur explanation

3+6+1 = 10

[JonathanSchneider]

Doh! Yeah, it seems I was getting tired as I typed that. You'll notice I has the 6 right and then changed it to a 7 mysteriously on the next line. Thanks for the catch!