Binary Numbers Arranged Puzzle



Write down the integers 1 through 10 in binary.
So, with leading zeros, we have:
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010

These binary integers can be arranged so that no one-bit in any integer
is in the same position as a one-bit in the integer's adjacent integer.
In other words, if b(m) and b(m+1) are adjacent elements of the
permutation
written in binary, then (b(m) AND b(m+1)) = 0 for 1 <= m <= n-1.
(If a logical AND is applied to each pair of adjacent terms, the result

is zero.)

So, for example, in an allowable permutation 1010 can be next to 0001,
but 1010 can not be next to 1000.

The puzzle is to determine by hand the number of allowable
permutations.
(Figuring this out is a not too difficult challenge, using reasoning
similar to
that used to solve an easy sudoku puzzle.)

As easy as it is to solve this puzzle, there is still room to make a
mistake. Perhaps someone should use a computer to check all 10!
permutations to count the number of allowable permutations, just to be
sure. (I may have the wrong answer myself.)

thanks,
Leroy Quet

.



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