Mmm Well there are 4 ways to get from E to F
Now going from F to G
If you horizontal first then there are 4 ways
if you go down1 then across there are 3 ways
if you go down 2 then across there ae 2 ways
if you go all the way down first then there is 1 way.
4+3+2+1=10 ways
If that is correct then there would be 4*10=40 ways to go from E to F to G
The ways to get from E to F = [1 + 3] ! / [1! * 3!] = 4!/3! = 4 ways
Ways to get from F to G = [3 + 2]! / [3! * 2! ] = 5! / 12 = 120 / 12 = 10 ways
So
4 ways x 10 ways = 40 ways to go from E to G through F
Mmm Well there are 4 ways to get from E to F
Now going from F to G
If you horizontal first then there are 4 ways
if you go down1 then across there are 3 ways
if you go down 2 then across there ae 2 ways
if you go all the way down first then there is 1 way.
4+3+2+1=10 ways
If that is correct then there would be 4*10=40 ways to go from E to F to G
I actually can't take much credit for this one.......the answer comes straight from a "formula" supplied by Nauseated....
For any R x C grid arrangement where R represents the rows and C represents the columns.....the number of distinct paths from one corner of the grid to a diagonally opposite corner is given by :
(R + C ) ! / [R! * C! ]
Minimum amount of steps E -> F -> G is 8, but 9 steps is required.
I.e. E -> F in 4 steps & F -> G in 5 steps, or E -> F in 5 steps & F -> G in 4 steps.
So is 40 ways to go from E to F to G in 8 steps, or did you somehow take that additional step into account?
(I don't have spare time to calculate, so cannot check this.. so just wondering)
Hi anon, thanks for getting involved - we like that.
However, the minimum number is 9. 5 across and 4 down
I don't know where you got 8 from? :)
Hi Chris,
Yes i thought that might be Nauseated's formula, It is nifty,
There are so many formuas for so many things i cannot remember them all.
I suppose i should try and organise a file for some of these new ones.