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