Just one question now...The rest I get just fine! If there are 10 identical yellow balls and 15 identical purple balls then how many distinguishable arrangements are possible if two yellows CANNOT be next to each other? ANSWER: 8008 My issue is not figuring out the answer, but the calculator equation that leads to it. This thing has been extremely annoying, but I'm "very" close to figuring out the equation (I have a TI-83+). Thanks!

SUM n=1..6 [ combinations(11;n) * combinations(5;n-1) ] its the third part that is bugging me... I understand the 11, and why the sum has six parts. But though the last part looks right and gives the right answer (phew) I've not yet grasped what it is representing in this particular problem. ========================================= [del]Here's a simpler formula that I have yet to figure out.[/del] Ans = combination(Purple+1;yellow) okay I've understood this one now. Place all the purples in a row which makes P+1 spaces for the yellows. Inserting each of the ten yellows one at a time into gaps is like drawing lots without replacement. That gives the number of permutations and we need to reduce it to combinations.

I've seen this exact problem before, but it was called "Balls and Bins" or "Bars and Stars". Basically, you have X amount of identical balls and Y amount of identical bins, with X>Y. You have to figure out how many combinations of balls into bins there are. Similar with Bars and Stars. X number of Stars, Y number of Bars. You have to place the Bars in between the Stars with no two Bars being adjacent. I like Bars and Stars better, because it's much easier to explain. Define the "Bars" to be the Yellow balls, and the "Stars" to be the Purple balls. You need to place the Bars in between the Stars. In this case, there are 16 spaces between the Stars (before and after also count). Code: _*_*_*_*_*_*_*_*_*_*_*_*_*_*_*_ Think about it like that. Now, considering that all the Bars are identical (as in, don't number them), and all the Stars are identical, find a way to put all the Bars into those 16 spaces. The formula is 16 choose 10 = 8008. (or, as NoPoke put it, Combination(16;10). His second explanation is correct) NoPoke's first answer might be very similar to this one, only he switched the Bars and the Stars definition. The 11 probably comes from the spaces in between the Yellow balls.

Hmmm...is this a Stat and Prob 2 problem? I ask because we're in finals/review in S&P1 right now, and we haven't covered anything exactly like this. Closest we've gotten is a problem whose solution involves the multiplication of two factorials.

Sort of. I learned it in a Computer Science theory class that focused heavily on aspects of probability and statistics. "Discrete Mathematical Structures" was the name of the class, and we worked on this somewhere in the vicinity of week 6 of the semester.