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