[Boron]

Originally posted by liga:

quote:

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Originally posted by Boron:

liga could you please share how you come to 6 , 15 and 20 times exactly ?

yes, the possible permutations between N objects of which n are in position A and (N-n) are in position B are

N! / ( n! * ( N-n)! )

so, we have N = 6

for 5+ n = 5 and N-n = 1 so 6! / (5! * 1!) = 6
for 4+ n = 4 and N-n = 2 so 6! / (4! * 2!) = 15
for 3+ n = 3 and N-n = 3 so 6! / (3! * 3!) = 20

bye bye
Liga

BTW. about the mod ... it seems great! I hope to have the possibility to play it soon!

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ok thnx . so you chose this formula
but i am unsure if you may only chose this formula or if you have to include something else .
because for 5+ of 1 path i think there should be 12 possible solutions :

AAAAAW , AAAAAE , AAAAAF , WWWWWA , WWWWWE , WWWWWF , EEEEEW , EEEEEA , EEEEEF , FFFFFA , FFFFFW , FFFFFE .

so 12 not 6 possible permutations right ?

i am unsure which formula(s) to use to get this outcome .
for 5+ it is easy to find it out because you can do the 12 permutations manually .
but for 3+ etc. you need a formula .

i forgot the permutations but i think you may not use your formula because this way you miss 6 possible permutations for 5+ example .

but i don't find a formula for this now so please tell me if i made a fault again or when you found a new one