UKMT Senior Challenge-XVIII
The diagram shows a square divided into six smaller squares labelled A, B, C, D, E and F. Two squares are considered to be adjacent if they have more than one point in common. The numbers 1, 2, 3, 4, 5 and 6 are to be placed in the smaller squares, one in each, so that no two adjacent squares contain numbers differing by 3. How many different arrangements are possible?
This problem is not original.This problem is part of this set .
The answer is 96.
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1 solution
The numbers 1, 2, 3, 4, 5 and 6 can be paired uniquely (1 with 4, 2 with 5, and 3 with 6)
So that the difference between the numbers in each pair is 3.
Any of the six numbers can be placed in position F but once that has been done, the partner of this number must be placed in position C as, otherwise, the square containing F would be adjacent to a square containing a number that differs by 3. Any of the remaining four numbers may be placed at position A , but then the number placed at B must be one of the two numbers from the pair that has so far been left unused.
Finally, the two unused numbers may be positioned at D and E , in either order.
Thus the number of possible arrangements is 6 × 4 × 3 × 2 which is 96.