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COUNTER CROSSES.





(Combination and Group Problems)
All that we need for this puzzle is nine counters, numbered 1, 2, 3, 4,
5, 6, 7, 8, and 9. It will be seen that in the illustration A these are
arranged so as to form a Greek cross, while in the case of B they form a
Latin cross. In both cases the reader will find that the sum of the
numbers in the upright of the cross is the same as the sum of the
numbers in the horizontal arm. It is quite easy to hit on such an
arrangement by trial, but the problem is to discover in exactly how many
different ways it may be done in each case. Remember that reversals and
reflections do not count as different. That is to say, if you turn this
page round you get four arrangements of the Greek cross, and if you turn
it round again in front of a mirror you will get four more. But these
eight are all regarded as one and the same. Now, how many different ways
are there in each case?
[Illustration:
(1) (2)
(2) (4) (5) (1) (6) (7)
(3) (4) (9) (5) (6) (3)
(7) (8)
A (8) B (9)
]


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