If we have a set like this { 1A ,2A ,2B, 3A, 3B, 3C}, how many non decreasing sequence is possible, such that number in left is less than number in right of length k? i.e, Length = 2 then the following sequence are possible: 1A 1A, 1A 2A, 1A 2B, 1A 3A, 1A 3B, 1A 3C, 2A 2B, 2B 2A, 2A 3A, ..... 3C 3C in total 25 combinations would be possible.
2026-04-09 03:34:21.1775705661
How many non decreasing sequence of length k is possible?
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Put X = 1, Y = 2, Z = 3 and compute for a few values of k. E.g. for k = 2, you have
XX = 1, XY = 2, XZ = 3, YY = 4, YZ = 6, ZZ = 9, giving your total of 25
Similarly, for k = 3, compute XXX, XXY, XXZ, XYY, XYZ, XZZ, YYY, YYZ,YZZ & ZZZ, and sum up.
Discern commonality between successive values of k, and work out through recursion.
You will get $f_n = 3f_{n-1} - 2f_{n-2} + 3^n$,
with $f_0 = 1, f_1 = 6$ as initial values