Chapter 33

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 × 2 × 3 × 4 = 24 ways. And as the four in the horizontal may also be changed in 24 ways for every arrangement on the other arm, we find that there are 24 × 24 = 576 variations for every form; therefore, as there are 18 forms, we get 18 × 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 × 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.

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.

The general solution to the problem is, in fact, this:

n C 2n ----- n + 1

where 2n equals the number of barrels. The symbol C, of course, implies that we have to find how many combinations, or selections, we can make of 2n things, taken n at a time.

280.--BUILDING THE TETRAHEDRON.

Take your constructed pyramid and hold it so that one stick only lies on the table. Now, four sticks must branch off from it in different directions--two at each end. Any one of the five sticks may be left out of this connection; therefore the four may be selected in 5 different ways. But these four matches may be placed in 24 different orders. And as any match may be joined at either of its ends, they may further be varied (after their situations are settled for any particular arrangement) in 16 different ways. In every arrangement the sixth stick may be added in 2 different ways. Now multiply these results together, and we get 5 × 24 × 16 × 2 = 3,840 as the exact number of ways in which the pyramid may be constructed. This method excludes all possibility of error.

A common cause of error is this. If you calculate your combinations by working upwards from a basic triangle lying on the table, you will get half the correct number of ways, because you overlook the fact that an equal number of pyramids may be built on that triangle downwards, so to speak, through the table. They are, in fact, reflections of the others, and examples from the two sets of pyramids cannot be set up to resemble one another--except under fourth dimensional conditions!

281.--PAINTING A PYRAMID.

It will be convenient to imagine that we are painting our pyramids on the flat cardboard, as in the diagrams, before folding up. Now, if we take any _four_ colours (say red, blue, green, and yellow), they may be applied in only 2 distinctive ways, as shown in Figs, 1 and 2. Any other way will only result in one of these when the pyramids are folded up. If we take any _three_ colours, they may be applied in the 3 ways shown in Figs. 3, 4, and 5. If we take any _two_ colours, they may be applied in the 3 ways shown in Figs. 6, 7, and 8. If we take any _single_ colour, it may obviously be applied in only 1 way. But four colours may be selected in 35 ways out of seven; three in 35 ways; two in 21 ways; and one colour in 7 ways. Therefore 35 applied in 2 ways = 70; 35 in 3 ways = 105; 21 in 3 ways = 63; and 7 in 1 way = 7. Consequently the pyramid may be painted in 245 different ways (70 + 105 + 63 + 7), using the seven colours of the solar spectrum in accordance with the conditions of the puzzle.

282.--THE ANTIQUARY'S CHAIN.

THE number of ways in which nine things may be arranged in a row without any restrictions is 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 = 362,880. But we are told that the two circular rings must never be together; therefore we must deduct the number of times that this would occur. The number is 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40,320 × 2 = 80,640, because if we consider the two circular links to be inseparably joined together they become as one link, and eight links are capable of 40,320 arrangements; but as these two links may always be put on in the orders AB or BA, we have to double this number, it being a question of arrangement and not of design. The deduction required reduces our total to 282,240. Then one of our links is of a peculiar form, like an 8. We have therefore the option of joining on either one end or the other on every occasion, so we must double the last result. This brings up our total to 564,480.

We now come to the point to which I directed the reader's attention--that every link may be put on in one of two ways. If we join the first finger and thumb of our left hand horizontally, and then link the first finger and thumb of the right hand, we see that the right thumb may be either above or below. But in the case of our chain we must remember that although that 8-shaped link has two independent _ends_ it is like every other link in having only two _sides_--that is, you cannot turn over one end without turning the other at the same time.

We will, for convenience, assume that each link has a black side and a side painted white. Now, if it were stipulated that (with the chain lying on the table, and every successive link falling over its predecessor in the same way, as in the diagram) only the white sides should be uppermost as in A, then the answer would be 564,480, as above--ignoring for the present all reversals of the completed chain. If, however, the first link were allowed to be placed either side up, then we could have either A or B, and the answer would be 2 × 564,480 = 1,128,960; if two links might be placed either way up, the answer would be 4 × 564,480; if three links, then 8 × 564,480, and so on. Since, therefore, every link may be placed either side up, the number will be 564,480 multiplied by 2^9, or by 512. This raises our total to 289,013,760.

But there is still one more point to be considered. We have not yet allowed for the fact that with any given arrangement three of the other arrangements may be obtained by simply turning the chain over through its entire length and by reversing the ends. Thus C is really the same as A, and if we turn this page upside down, then A and C give two other arrangements that are still really identical. Thus to get the correct answer to the puzzle we must divide our last total by 4, when we find that there are just 72,253,440 different ways in which the smith might have put those links together. In other words, if the nine links had originally formed a piece of chain, and it was known that the two circular links were separated, then it would be 72,253,439 chances to 1 that the smith would not have put the links together again precisely as they were arranged before!

283.--THE FIFTEEN DOMINOES.

The reader may have noticed that at each end of the line I give is a four, so that, if we like, we can form a ring instead of a line. It can easily be proved that this must always be so. Every line arrangement will make a circular arrangement if we like to join the ends. Now, curious as it may at first appear, the following diagram exactly represents the conditions when we leave the doubles out of the question and devote our attention to forming circular arrangements. Each number, or half domino, is in line with every other number, so that if we start at any one of the five numbers and go over all the lines of the pentagon once and once only we shall come back to the starting place, and the order of our route will give us one of the circular arrangements for the ten dominoes. Take your pencil and follow out the following route, starting at the 4: 41304210234. You have been over all the lines once only, and by repeating all these figures in this way, 41--13--30--04--42--21--10--02--23--34, you get an arrangement of the dominoes (without the doubles) which will be perfectly clear. Take other routes and you will get other arrangements. If, therefore, we can ascertain just how many of these circular routes are obtainable from the pentagon, then the rest is very easy.

Well, the number of different circular routes over the pentagon is 264. How I arrive at these figures I will not at present explain, because it would take a lot of space. The dominoes may, therefore, be arranged in a circle in just 264 different ways, leaving out the doubles. Now, in any one of these circles the five doubles may be inserted in 2^5 = 32 different ways. Therefore when we include the doubles there are 264 × 32 = 8,448 different circular arrangements. But each of those circles may be broken (so as to form our straight line) in any one of 15 different places. Consequently, 8,448 × 15 gives 126,720 different ways as the correct answer to the puzzle.

I purposely refrained from asking the reader to discover in just how many different ways the full set of twenty-eight dominoes may be arranged in a straight line in accordance with the ordinary rules of the game, left to right and right to left of any arrangement counting as different ways. It is an exceedingly difficult problem, but the correct answer is 7,959,229,931,520 ways. The method of solving is very complex.

284.--THE CROSS TARGET.

Twenty-one different squares may be selected. Of these nine will be of the size shown by the four A's in the diagram, four of the size shown by the B's, four of the size shown by the C's, two of the size shown by the D's, and two of the size indicated by the upper single A, the upper single E, the lower single C, and the EB. It is an interesting fact that you cannot form any one of these twenty-one squares without using at least one of the six circles marked E.

285.--THE FOUR POSTAGE STAMPS.

Referring to the original diagram, the four stamps may be given in the shape 1, 2, 3, 4, in three ways; in the shape 1, 2, 5, 6, in six ways; in the shape 1, 2, 3, 5, or 1, 2, 3, 7, or 1, 5, 6, 7, or 3, 5, 6, 7, in twenty-eight ways; in shape 1, 2, 3, 6, or 2, 5, 6, 7, in fourteen ways; in shape 1, 2, 6, 7, or 2, 3, 5, 6, or 1, 5, 6, 10, or 2, 5, 6, 9, in fourteen ways. Thus there are sixty-five ways in all.

286.--PAINTING THE DIE.

The 1 can be marked on any one of six different sides. For every side occupied by 1 we have a selection of four sides for the 2. For every situation of the 2 we have two places for the 3. (The 6, 5, and 4 need not be considered, as their positions are determined by the 1, 2, and 3.) Therefore 6, 4, and 2 multiplied together make 48 different ways--the correct answer.

287.--AN ACROSTIC PUZZLE.

There are twenty-six letters in the alphabet, giving 325 different pairs. Every one of these pairs may be reversed, making 650 ways. But every initial letter may be repeated as the final, producing 26 other ways. The total is therefore 676 different pairs. In other words, the answer is the square of the number of letters in the alphabet.

288.--CHEQUERED BOARD DIVISIONS.

There are 255 different ways of cutting the board into two pieces of exactly the same size and shape. Every way must involve one of the five cuts shown in Diagrams A, B, C, D, and E. To avoid repetitions by reversal and reflection, we need only consider cuts that enter at the points a, b, and c. But the exit must always be at a point in a straight line from the entry through the centre. This is the most important condition to remember. In case B you cannot enter at a, or you will get the cut provided for in E. Similarly in C or D, you must not enter the key-line in the same direction as itself, or you will get A or B. If you are working on A or C and entering at a, you must consider joins at one end only of the key-line, or you will get repetitions. In other cases you must consider joins at both ends of the key; but after leaving a in case D, turn always either to right or left--use one direction only. Figs. 1 and 2 are examples under A; 3 and 4 are examples under B; 5 and 6 come under C;

and 7 is a pretty example of D. Of course, E is a peculiar type, and obviously admits of only one way of cutting, for you clearly cannot enter at b or c.

Here is a table of the results:--

a b c Ways. A = 8 + 17 + 21 = 46 B = 0 + 17 + 21 = 38 C = 15 + 31 + 39 = 85 D = 17 + 29 + 39 = 85 E = 1 + 0 + 0 = 1 -- -- -- --- 41 94 120 255

I have not attempted the task of enumerating the ways of dividing a board 8 × 8--that is, an ordinary chessboard. Whatever the method adopted, the solution would entail considerable labour.

289.--LIONS AND CROWNS.

Here is the solution. It will be seen that each of the four pieces (after making the cuts along the thick lines) is of exactly the same size and shape, and that each piece contains a lion and a crown. Two of the pieces are shaded so as to make the solution quite clear to the eye.

290.--BOARDS WITH AN ODD NUMBER OF SQUARES.

There are fifteen different ways of cutting the 5 × 5 board (with the central square removed) into two pieces of the same size and shape. Limitations of space will not allow me to give diagrams of all these, but I will enable the reader to draw them all out for himself without the slightest difficulty. At whatever point on the edge your cut enters, it must always end at a point on the edge, exactly opposite in a line through the centre of the square. Thus, if you enter at point 1 (see Fig. 1) at the top, you must leave at point 1 at the bottom. Now, 1 and 2 are the only two really different points of entry; if we use any others they will simply produce similar solutions. The directions of the cuts in the following fifteen

solutions are indicated by the numbers on the diagram. The duplication of the numbers can lead to no confusion, since every successive number is contiguous to the previous one. But whichever direction you take from the top downwards you must repeat from the bottom upwards, one direction being an exact reflection of the other.

1, 4, 8. 1, 4, 3, 7, 8. 1, 4, 3, 7, 10, 9. 1, 4, 3, 7, 10, 6, 5, 9. 1, 4, 5, 9. 1, 4, 5, 6, 10, 9. 1, 4, 5, 6, 10, 7, 8. 2, 3, 4, 8. 2, 3, 4, 5, 9. 2, 3, 4, 5, 6, 10, 9. 2, 3, 4, 5, 6, 10, 7, 8. 2, 3, 7, 8. 2, 3, 7, 10, 9. 2, 3, 7, 10, 6, 5, 9. 2, 3, 7, 10, 6, 5, 4, 8.

It will be seen that the fourth direction (1, 4, 3, 7, 10, 6, 5, 9) produces the solution shown in Fig. 2. The thirteenth produces the solution given in propounding the puzzle, where the cut entered at the side instead of at the top. The pieces, however, will be of the same shape if turned over, which, as it was stated in the conditions, would not constitute a different solution.

291.--THE GRAND LAMA'S PROBLEM.

The method of dividing the chessboard so that each of the four parts shall be of exactly the same size and shape, and contain one of the gems, is shown in the diagram. The method of shading the squares is adopted to make the shape of the pieces clear to the eye. Two of the pieces are shaded and two left white.

The reader may find it interesting to compare this puzzle with that of the "Weaver" (No. 14, _Canterbury Puzzles_).

292.--THE ABBOT'S WINDOW.

THE man who was "learned in strange mysteries" pointed out to Father John that the orders of the Lord Abbot of St. Edmondsbury might be easily carried out by blocking up twelve of the lights in the window as shown by the dark squares in the following sketch:--

Father John held that the four corners should also be darkened, but the sage explained that it was desired to obstruct no more light than was absolutely necessary, and he said, anticipating Lord Dundreary, "A single pane can no more be in a _line_ with itself than one bird can go into a corner and flock in solitude. The Abbot's condition was that no diagonal _lines_ should contain an odd number of lights."

Now, when the holy man saw what had been done he was well pleased, and said, "Truly, Father John, thou art a man of deep wisdom, in that thou hast done that which seemed impossible, and yet withal adorned our window with a device of the cross of St. Andrew, whose name I received from my godfathers and godmothers." Thereafter he slept well and arose refreshed. The window might be seen intact to-day in the monastery of St. Edmondsbury, if it existed, which, alas! the window does not.

293.--THE CHINESE CHESSBOARD.

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