There are two variations I would like to discuss about this problem that I have encountered and haven't been able to solve previously.↵
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1) Shuffling is introduced↵
i.e when order/arrangement matters.Like sSuppose wWe have a1 + a2 + a3 = 4 then one of the solutions is (a1,a2,a3) = (1,1,2) , but this should not be counted as 1 arrangement but instead as (1+1+2)! / 2! arrangements because the order of the arrangement matters i.e (a1,a2,a3,a3) is different from (a1,a3,a2,a3). Is there an efficient way to count this ? ↵
infinte (or >=N) flags of each of r colors. We have to find an arrangement of N flags using these flags . How many ways can this be done. (So a1 + a2 + .... ar = N where ai is number of flags of color i but since this is an arrangement, shuffling within the N flags is possible).↵
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2) No shuffling but similar sets should be counted only once. ↵
i.e (a1,a2,a3) = (1,1,2) is same as (a1,a2,a3) = (2,1,1)↵
↵
Also, what is a good blog/site/resource for intermediate-hard counting/combinatorics problems (with editorials/theory).
↵
1) Shuffling is introduced↵
i.e when order/arrangement matters.
↵
2) No shuffling but similar sets should be counted only once. ↵
i.e (a1,a2,a3) = (1,1,2) is same as (a1,a2,a3) = (2,1,1)↵
↵
Also, what is a good blog/site/resource for intermediate-hard counting/combinatorics problems (with editorials/theory).
