24 ways for every arrangement on the other arm, we find that there are 24 x 24 = 576 variations for every form; therefore, as there are 18 forms, we get 18 x 576 = 10,368 ways.  But this will include half the four reversals and half the four reflections that we barred, so we must divide this by 4 to obtain the correct answer to the Greek Cross, which is thus 2,592 different ways.  The division is by 4 and not by 8, because we provided against half the reversals and reflections by always reserving one number for the upright and the other for the horizontal.

In the case of the Latin Cross, it is obvious that we have to deal with the same 18 forms of pairing.  The total number of different ways in this case is the full number, 18 x 576.  Owing to the fact that the upper and lower arms are unequal in length, permutations will repeat by reflection, but not by reversal, for we cannot reverse.  Therefore this fact only entails division by 2.  But in every pair we may exchange the figures in the upright with those in the horizontal (which we could not do in the case of the Greek Cross, as the arms are there all alike); consequently we must multiply by 2.  This multiplication by 2 and division by 2 cancel one another.  Hence 10,368 is here the correct answer.

278.—­A DORMITORY PUZZLE.

[Illustration: 

         MON.  TUES.  WED.
    +—–­+—–­+—–­+ +—–­+—–­+—–­+ +—–­+—–­+—–­+
    | 1 | 2 | 1 | | 1 | 3 | 1 | | 1 | 4 | 1 |
    +—–­+—–­+—–­+ +—–­+—–­+—–­+ +—–­+—–­+—–­+
    | 2 | | 2 | | 1 | | 1 | | 1 | | 1 |
    +—–­+—–­+—–­+ +—–­+—–­+—–­+ +—–­+—–­+—–­+
    | 1 | 22| 1 | | 3 | 19| 3 | | 4 | 16| 4 |
    +—–­+—–­+—–­+ +—–­+—–­+—–­+ +—–­+—–­+—–­+

    THURS.  FRI.  SAT.

+—–­+—–­+—–­+ +—–­+—–­+—–­+ +—–­+—–­+—–­+
| 1 | 5 | 1 | | 2 | 6 | 2 | | 4 | 4 | 4 |
+—–­+—–­+—–­+ +—–­+—–­+—–­+ +—–­+—–­+—–­+
| 2 | | 2 | | 1 | | 1 | | 4 | | 4 |
+—–­+—–­+—–­+ +—–­+—–­+—–­+ +—–­+—–­+—–­+
| 4 | 13| 4 | | 7 | 6 | 7 | | 4 | 4 | 4 |
+—–­+—–­+—–­+ +—–­+—–­+—–­+ +—–­+—–­+—–­+

]

Arrange the nuns from day to day as shown in the six diagrams.  The smallest possible number of nuns would be thirty-two, and the arrangements on the last three days admit of variation.

279.—­THE BARRELS OF BALSAM.

This is quite easy to solve for any number of barrels—­if you know how.  This is the way to do it.  There are five barrels in each row Multiply the numbers 1, 2, 3, 4, 5 together; and also multiply 6, 7, 8, 9, 10 together.  Divide one result by the other, and we get the number of different combinations or selections of ten things taken five at a time.  This is here 252.  Now, if we divide this by 6 (1 more than the number in the row) we get 42, which is the correct answer to the puzzle, for there are 42 different ways of arranging the barrels.  Try this method of solution in the case of six barrels, three in each row, and you will find the answer is 5 ways.  If you check this by trial, you will discover the five arrangements with 123, 124, 125, 134, 135 respectively in the top row, and you will find no others.