An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

Choose any three by three square of dates on a calendar page...

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1 Step, 2 Step Poster

233 ways

Why?

Method 1 - using shorter staircases

Staircase with 1 step: 1 way

Staircase with 2 steps: 1, 1 or 2.

Staircase with 3 steps: 1+staircase with 2 steps $\rightarrow$ 2 ways, or 2+staircase with 1 step $\rightarrow$ 1 way, total 3 ways

Staircase with 4 steps: 1+staircase with 3 steps $\rightarrow$ 3 ways, or 2+staircase with 2 steps $\rightarrow$ 2 ways, total 5 ways

Staircase with 5 steps: 1+staircase with 4 steps $\rightarrow$ 5 ways, or 2+staircase with 3 steps $\rightarrow$ 3 ways, total 8 ways

Staircase with 6 steps: 8 + 5 = 13 ways

Staircase with 7 steps: 13 + 8 = 21 ways

Staircase with 6 steps: 21 + 13 = 34 ways

Staircase with 6 steps: 34 + 21 = 55 ways

Staircase with 6 steps: 55 + 34 = 89 ways

Staircase with 6 steps: 89 + 55 = 144 ways

Staircase with 6 steps: 144 + 89 = 233 ways

Method 2 - using binomial coefficients

Numbers of 1s and 2s that add up to 12

1 step	2 steps	Number of jumps	Number of possible ways
12	0	12	1 way - 1, 1, 1, ..., 1
10	1	11	11 ways - the 2 could be any one of the 11 jumps
8	2	10	10 options for the 'first' 2, 9 options for the 'second' 2 - but we have to divide by 2 afterwards because the two 2s are interchangeable, there is no 'first' and 'second' 10$\times$9$\div$2 = 45
6	3	9	9 options for the 'first' 2, 8 for the 'second' and 7 for the 'third' - but now each option has been counted 3$\times$2$\times$1 times 9$\times$8$\times$7$\div$6 = 84
4	4	8	8$\times$7$\times$6$\times$5$\div$(4$\times$3$\times$2$\times$1 = 70
2	5	7	Now use the options for the two 1s, to make it easier 7$\times$6$\div$2 = 21
0	6	6	1 way - 2, 2, 2, ...

Total number of ways: 1 + 11 + 45 + 84 + 70 + 21 + 1 = 233