I have seen the problem in this forum about 6 number 1,1,2,2,3,3 to be arranged in a line with no same adjacent numbers. The solution is 30. But what happens if there are 8 number 1,1,2,2,3,3,4,4? Are there any general rules and quick ways to solve? I have tried to 8!/(2!2!2!2!)- 4 pairs -3 pairs-2pairs - 1pair, but the computations are too cumbersome....any good ways to solve? please help
on second thoughts the solution is not right:
the formula can eb expanded when N =4 to:
8!/2^4-4C1.6!/2^3+...
the first term describes all possible combinations
but the 4C1.6!/2^3 only takes in the first 2 positions being a pair, and all possible combinations of the remaining 6 places.
similarly 3rd term only describes the first 4 places being 2 pairs, and all possible combinations of the remaining 4 places...
this does not account for x pair pair xx..
I think we have to work again.
thx for your helpful input..
rgds