277.--COUNTER CROSSES.

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)

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