13 steps

[uksjn]

ok so if there is thirteen steps in a staircase how many variants are there if you can only step 1 or 2 steps at any 1 time

thankyou for doing this it has frustrated me for a week or two :lol:

 

[Deleted member 4993]

ok so if there is thirteen steps in a staircase how many variants are there if you can only step 1 or 2 steps at any 1 time

thankyou for doing this it has frustrated me for a week or two :lol:

Please show us some of the steps you have tried.

 

[pka]

ok so if there is thirteen steps in a staircase how many variants are there if you can only step 1 or 2 steps at any 1 time

Here is the exact answer. \(\displaystyle \sum\limits_{k - 0}^6 {\frac{{(13 - k)!}}{{k!(13 - 2k)!}} = 377}\)

Here is a way to do it recursively for any number of N steps.
If \(\displaystyle N=1\) then \(\displaystyle D_1=1\).
If \(\displaystyle N=2\) then \(\displaystyle D_1=2\).

Now suppose that \(\displaystyle N\ge 3\) then \(\displaystyle D_N=D_{N-1}+D_{N-2}\).
Why? Because, from the (N-2)[sup:xq0tb3gv]th[/sup:xq0tb3gv] step I can get to the N[sup:xq0tb3gv]th[/sup:xq0tb3gv] step by taking two more steps.
And from the (N-1)[sup:xq0tb3gv]th[/sup:xq0tb3gv] step I can get to the N[sup:xq0tb3gv]th[/sup:xq0tb3gv] step by taking one more step.

 

[uksjn]

i have tried to work it out up to 5 steps and seeing if there is any pattern

 

[pka]

i have tried to work it out up to 5 steps and seeing if there is any pattern

\(\displaystyle \begin{array}{rcr} 1 &\vline & 1 \\ 2 &\vline & 2 \\ 3 &\vline & 3 \\ 4 &\vline & 5 \\ 5 &\vline & 8 \\ 6 &\vline & {13} \\ 7 &\vline & {21} \\ 8 &\vline & {34} \\ 9 &\vline & {55} \\ {10} &\vline & {89} \\ {11} &\vline & {144} \\ {12} &\vline & {233} \\ {13} &\vline & {377} \\ \end{array}\)

Have you ever heard of Fibonacci numbers?

 

[uksjn]

thankyou i am so happy that you done this for me