Question 1207714: In how many ways can the numbers 1, 2, 3, 4, 5, and 6 be arranged in a row, so that the product of any two adjacent numbers is at least 4?
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!
The only possible adjacent positive integer pairs with a
product less than 4 are "12", "21", "13", and "31".
So we start with 6!=720 arrangements of 1,2,3,4,5,6.
Then we subtract the 5!=120 arrangements of (12),3,4,5,6 and
the 5!=120 arrangements of (21),3,4,5,6.
Then we subtract the 5!=120 arrangements of (13),2,4,5,6 and
the 5!=120 arrangements of (31),2,4,5,6.
But we have subtracted out those containing "213" twice
and those containing "312" twice. So we need to add back in
the 4!=24 arrangements of (213),4,5,6, and the 4!=24
arrangements of (312),4,5,6, once each.
So the answer is
6! - 4x5! + 2x4! = 288.
Edwin
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