Binary Strings

For the ternary string:

012xx
x012x
xx012

So you have three cases, each case has nine possible permutations.

The first problem is a standard probability problem. It is the same question as "you have three red balls and five blue balls. How many permutations can you make?" The answer is 8!/(5! 3!)

The second problem can also be simplified into a probability problem, but you don't use factorial because the values that can go into each slot are not dependent on the previous values used like the above problem. You know three of your five slots are taken by 012. You need to know how many permutations you can make with the remaining two positions given that each one could be one of three options. For each position of 012, there would be 3^2 total permutations. There are three different places you could place the 012, so multiply that by three again for a total of 27. Here is an example:

With 012 up front
012 00
012 01
012 02
012 10
012 11
etc.

Next with 012 in the middle
0 012 0
0 012 1
0 012 2
1 012 0
1 012 1
etc.

You can do the same with 012 on the end.

thanks i think i understand that now.

Ok heres another one along the same line, for the fibonacci series where f0=f1=1 and fn+1=fn+fn-1 prove fn = the sum of (n-r)!/(r!(n-2r)!) (from r=0 to n) does anyone have any starting points for this i havent a clue where to begin!

ok sorry ive got that now had a brain wave

but can anyone tell me how to show that the sum of fn/2^n = 4