|Mar14-10, 11:14 AM||#18|
Permutations and Combinations Question | Unlimited balls of 4 varieties |
Hey everyone, I find this thread very interesting, but I am having trouble follwing the idea. I hope someone will be kind enough to expound the reasoning for me.
I figure the answer would simply be 49, as there are four ways to choose the first colour, four the second, and so forth. This is clearly not the case, but I am having difficult understanding why.
A comprehensive explanation is not necessary, but any pointers would be helpful.
|Mar14-10, 11:43 AM||#19|
Short pointer: order doesn't matter.
Longer pointer: blue, blue, green, red, ..., red is to be considered the same as blue, green, blue, red, ..., red. In mathematical language, we're considering multisets.
You can pick the 9 balls as follows: a blue ones, b green ones, c red ones, d white ones.
The question is, how many different pairs (a,b,c,d) are there such that a+b+c+d=9.
|Mar14-10, 01:10 PM||#20|
Many thanks for the response and for clearing up that confusion.
|balls, combinations, number of ways, permutations|
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