a rabbit can climb one or two steps at a time on a ten step staircase, but he never goes back down. In how many different ways can the rabbit climb a flight of ten steps
5 2's -- one way
4 2's 2 1's ---- number of ways = 6!/(4!2!) = 15
(e.g. 222211, 11222, 212122, etc )
3 2's 4 1's ---- number of ways = 7!/(4!3!) = 35
2 2's 6 1's ---- number of ways = 8!/(2!6!) = 28
1 2 8 1's ---- number of ways = 9!/(1!8!) = 9
all 1's ---- one way
total 1+15+35+28+9+1 = 89
Wow for grade 8!!!
89
I'm lost... what does 7i(backwords)(4i3i) means?
What does the i facing down stands for?
To find the number of different ways the rabbit can climb a flight of ten steps, we can use a dynamic programming approach.
Let's consider the number of ways to climb each step individually. The rabbit can either climb one step or two steps at a time, but never goes back down. This means that the number of ways to reach step n is the sum of the number of ways to reach step n-1 and the number of ways to reach step n-2.
Using this logic, we can start computing the number of ways to reach each step starting from the base cases: the number of ways to reach the 1st step is 1 (by climbing one step) and the number of ways to reach the 2nd step is 2 (by climbing either one step twice or two steps once).
We can then build up the Fibonacci sequence to find the number of ways to reach the subsequent steps:
Step 1: 1
Step 2: 2
Step 3: 3 (1 + 2)
Step 4: 5 (2 + 3)
Step 5: 8 (3 + 5)
Step 6: 13 (5 + 8)
Step 7: 21 (8 + 13)
Step 8: 34 (13 + 21)
Step 9: 55 (21 + 34)
Step 10: 89 (34 + 55)
Therefore, the rabbit can climb a flight of ten steps in 89 different ways.
In general, if you want to find the number of ways to climb an n-step staircase using steps of 1 or 2, you can use the formula for the nth Fibonacci number, which is:
Fn = Fn-1 + Fn-2
where F1 = 1 and F2 = 2.