+ 55 = 235 ; with 3, the same as with 4, + 52 = 287; with 2, the same as with 3, + 45 = 332; with 1, the same as with 2, + 34 = 366, and with nought in the top left-hand corner the number of solutions will be found to be 10 + 27 + 40 + 49 + 54 + 55 + 52 + 45 + 34 + 19 = 385.  As there is no other number to be placed in the top left-hand corner, we have now only to add these totals together thus, 10 + 37 + 77 + 126 + 180 + 235 + 287 + 332 + 366 + 385 = 2,035.  We therefore find that the total number of ways in which tenants may occupy some or all of the eight villas so that there shall be always nine persons living along each side of the square is 2,035.  Of course, this method must obviously cover all the reversals and reflections, since each corner in turn is occupied by every number in all possible combinations with the other two corners that are in line with it.

[Illustration: 

       A B C D E
    +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+
    |9| |0| |8| |0| |8| |1| |8| |0| |8| |1|
    +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+
    | |*| | | |*| | | |*| | | |*| | | |*| |
    +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+
    |0| | | |0| | | |1| | | |1| | | |0| | |
    +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+

]

Here is a general formula for solving the puzzle:  (n squared + 3n + 2)(n squared + 3n + 3)/6.  Whatever may be the stipulated number of residents along each of the sides (which number is represented by n), the total number of different arrangements may be thus ascertained.  In our particular case the number of residents was nine.  Therefore (81 + 27 + 2) x (81 + 27 + 3) and the product, divided by 6, gives 2,035.  If the number of residents had been 0, 1, 2, 3, 4, 5, 6, 7, or 8, the total arrangements would be 1, 7, 26, 70, 155, 301, 532, 876, or 1,365 respectively.

277.—­COUNTER CROSSES.

Let us first deal with the Greek Cross.  There are just eighteen forms in which the numbers may be paired for the two arms.  Here they are:—­

    12978 13968 14958
    34956 24957 23967

    23958 13769 14759
    14967 24758 23768

    12589 23759 13579
    34567 14768 24568

    14569 23569 14379
    23578 14578 25368

    15369 24369 23189
    24378 15378 45167

    24179 25169 34169
    35168 34178 25178

Of course, the number in the middle is common to both arms.  The first pair is the one I gave as an example.  I will suppose that we have written out all these crosses, always placing the first row of a pair in the upright and the second row in the horizontal arm.  Now, if we leave the central figure fixed, there are 24 ways in which the numbers in the upright may be varied, for the four counters may be changed in 1 x 2 x 3 x 4 = 24 ways.  And as the four in the horizontal may also be changed in