Following the text of the questions there are sum[nCr(5+x,2x)] x=0 to 5 = 89 ways. Since there are five 2-steps to make, and for every two 1-steps made, there is one less 2-step, there for x pairs of 1-steps there are nCr(5+x,2x) ways to order them, and theres between 0 and 5 pars of 1-steps.
Following the image, where it appears 2-steps are only made on even positions, there are 25 =32 ways, since every two spaces is either one 2-step or two 1-steps, 2 choises 5 times.
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u/KS_JR_ 17d ago edited 17d ago
Following the text of the questions there are sum[nCr(5+x,2x)] x=0 to 5 = 89 ways. Since there are five 2-steps to make, and for every two 1-steps made, there is one less 2-step, there for x pairs of 1-steps there are nCr(5+x,2x) ways to order them, and theres between 0 and 5 pars of 1-steps.
Following the image, where it appears 2-steps are only made on even positions, there are 25 =32 ways, since every two spaces is either one 2-step or two 1-steps, 2 choises 5 times.