Part 35

We thus have twenty-six good words and one doubtful, obtained under the required conditions, and I do not think it will be easy to improve on this answer. Of course we are not bound by dictionaries but by common usage. If we went by the dictionary only in a case of this kind, we should find ourselves involved in prefixes, contractions, and such absurdities as I.O.U., which Nuttall actually gives as a word.

272.--THE NINE SCHOOLBOYS.

The boys can walk out as follows:--

1st Day. 2nd Day. 3rd Day. A B C B F H F A G D E F E I A I D B G H I C G D H C E

4th Day. 5th Day. 6th Day. A D H G B I D C A B E G C F D E H B F I C H A E I G F

Every boy will then have walked by the side of every other boy once and once only.

Dealing with the problem generally, 12n+9 boys may walk out in triplets under the conditions on 9n+6 days, where n may be nought or any integer. Every possible pair will occur once. Call the number of boys m. Then every boy will pair m-1 times, of which (m-1)/4 times he will be in the middle of a triplet and (m-1)/2 times on the outside. Thus, if we refer to the solution above, we find that every boy is in the middle twice (making 4 pairs) and four times on the outside (making the remaining 4 pairs of his 8). The reader may now like to try his hand at solving the two next cases of 21 boys on 15 days, and 33 boys on 24 days. It is, perhaps, interesting to note that a school of 489 boys could thus walk out daily in one leap year, but it would take 731 girls (referred to in the solution to No. 269) to perform their particular feat by a daily walk in a year of 365 days.

273.--THE ROUND TABLE.

The history of this problem will be found in _The Canterbury Puzzles_ (No. 90). Since the publication of that book in 1907, so far as I know, nobody has succeeded in solving the case for that unlucky number of persons, 13, seated at a table on 66 occasions. A solution is possible for any number of persons, and I have recorded schedules for every number up to 25 persons inclusive and for 33. But as I know a good many mathematicians are still considering the case of 13, I will not at this stage rob them of the pleasure of solving it by showing the answer. But I will now display the solutions for all the cases up to 12 persons inclusive. Some of these solutions are now published for the first time, and they may afford useful clues to investigators.

The solution for the case of 3 persons seated on 1 occasion needs no remark.

A solution for the case of 4 persons on 3 occasions is as follows:--

1 2 3 4 1 3 4 2 1 4 2 3

Each line represents the order for a sitting, and the person represented by the last number in a line must, of course, be regarded as sitting next to the first person in the same line, when placed at the round table.

The case of 5 persons on 6 occasions may be solved as follows:--

1 2 3 4 5 1 2 4 5 3 1 2 5 3 4 --------- 1 3 2 5 4 1 4 2 3 5 1 5 2 4 3

The case for 6 persons on 10 occasions is solved thus:--

1 2 3 6 4 5 1 3 4 2 5 6 1 4 5 3 6 2 1 5 6 4 2 3 1 6 2 5 3 4 ----------- 1 2 4 5 6 3 1 3 5 6 2 4 1 4 6 2 3 5 1 5 2 3 4 6 1 6 3 4 5 2

It will now no longer be necessary to give the solutions in full, for reasons that I will explain. It will be seen in the examples above that the 1 (and, in the case of 5 persons, also the 2) is repeated down the column. Such a number I call a "repeater." The other numbers descend in cyclical order. Thus, for 6 persons we get the cycle, 2, 3, 4, 5, 6, 2, and so on, in every column. So it is only necessary to give the two lines 1 2 3 6 4 5 and 1 2 4 5 6 3, and denote the cycle and repeaters, to enable any one to write out the full solution straight away. The reader may wonder why I do not start the last solution with the numbers in their natural order, 1 2 3 4 5 6. If I did so the numbers in the descending cycle would not be in their natural order, and it is more convenient to have a regular cycle than to consider the order in the first line.

The difficult case of 7 persons on 15 occasions is solved as follows, and was given by me in _The Canterbury Puzzles_:--

1 2 3 4 5 7 6 1 6 2 7 5 3 4 1 3 5 2 6 7 4 1 5 7 4 3 6 2 1 5 2 7 3 4 6

In this case the 1 is a repeater, and there are _two_ separate cycles, 2, 3, 4, 2, and 5, 6, 7, 5. We thus get five groups of three lines each, for a fourth line in any group will merely repeat the first line.

A solution for 8 persons on 21 occasions is as follows:--

1 8 6 3 4 5 2 7 1 8 4 5 7 2 3 6 1 8 2 7 3 6 4 5

The 1 is here a repeater, and the cycle 2, 3, 4, 5, 6, 7, 8. Every one of the 3 groups will give 7 lines.

Here is my solution for 9 persons on 28 occasions:--

2 1 9 7 4 5 6 3 8 2 9 5 1 6 8 3 4 7 2 9 3 1 8 4 7 5 6 2 9 1 5 6 4 7 8 3

There are here two repeaters, 1 and 2, and the cycle is 3, 4, 5, 6, 7, 8, 9. We thus get 4 groups of 7 lines each.

The case of 10 persons on 36 occasions is solved as follows:--

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

The repeater is 1, and the cycle, 2, 3, 4, 5, 6, 7, 8, 9, 10. We here have 4 groups of 9 lines each.

My solution for 11 persons on 45 occasions is as follows:--

2 11 9 4 7 6 5 1 8 3 10 2 1 11 7 6 3 10 8 5 4 9 2 11 10 3 9 4 8 5 1 7 6 2 11 5 8 1 3 10 6 7 9 4 2 11 1 10 3 4 9 6 7 5 8

There are two repeaters, 1 and 2, and the cycle is, 3, 4, 5,... 11. We thus get 5 groups of 9 lines each.

The case of 12 persons on 55 occasions is solved thus:--

1 2 3 12 4 11 5 10 6 9 7 8 1 2 4 11 6 9 8 7 10 5 12 3 1 2 5 10 8 7 11 4 3 12 6 9 1 2 6 9 10 5 3 12 7 8 11 4 1 2 7 8 12 3 6 9 11 4 5 10

Here 1 is a repeater, and the cycle is 2, 3, 4, 5,... 12. We thus get 5 groups of 11 lines each.

274.--THE MOUSE-TRAP PUZZLE.

If we interchange cards 6 and 13 and begin our count at 14, we may take up all the twenty-one cards--that is, make twenty-one "catches"--in the following order: 6, 8, 13, 2, 10, 1, 11, 4, 14, 3, 5, 7, 21, 12, 15, 20, 9, 16, 18, 17, 19. We may also exchange 10 and 14 and start at 16, or exchange 6 and 8 and start at 19.

275.--THE SIXTEEN SHEEP.

The six diagrams on next page show solutions for the cases where we replace 2, 3, 4, 5, 6, and 7 hurdles. The dark lines indicate the hurdles that have been replaced. There are, of course, other ways of making the removals.

276.--THE EIGHT VILLAS.

There are several ways of solving the puzzle, but there is very little difference between them. The solver should, however, first of all bear in mind that in making his calculations he need only consider the four villas that stand at the corners, because the intermediate villas can never vary when the corners are known. One way is to place the numbers nought to 9 one at a time in the top left-hand corner, and then consider each case in turn.

Now, if we place 9 in the corner as shown in the Diagram A, two of the corners cannot be occupied, while the corner that is diagonally opposite may be filled by 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 persons. We thus see that there are 10

[Illustration:

+---+---+ +-+-----+ +---+---+ |O OHO O| |OHO O O| |O OHO O| | H | | + | | +=+ | |O OHO O| |OHO O O| |O OHOHO| +-+ +-+-+ +-+-----+ +---+ + | |O|O O|O| |O|O O O| |O O O|O| | +---+ | | +-+-+ | | +-+ | |O O O O| |O O OHO| |O O|O O| +-------+ +-------+ +-------+ 2 3 4

+-----+-+ +-+-----+ +-------+ |O O OHO| |OHO O O| |O O O O| | +=+ | | +=+ | | +=+=+=+ |O OHO O| |OHOHO O| |OHOHO O| | +-+-+ + | + +-+ | + + + | |O|O O|O| |O|O O|O| |O|OHO O| +=+ +=+ | + +=+ +=+ + | |O O O O| |OHO O O| |O O|O O| +-------+ +-+-----+ +---+---+ 5 6 7 THE SIXTEEN SHEEP

]

solutions with a 9 in the corner. If, however, we substitute 8, the two corners in the same row and column may contain 0, 0, or 1, 1, or 0, 1, or 1, 0. In the case of B, ten different selections may be made for the fourth corner; but in each of the cases C, D, and E, only nine selections are possible, because we cannot use the 9. Therefore with 8 in the top left-hand corner there are 10 + (3 × 9) = 37 different solutions. If we then try 7 in the corner, the result will be 10 + 27 + 40, or 77 solutions. With 6 we get 10 + 27 + 40 + 49 = 126; with 5, 10 + 27 + 40 + 49 + 54 = 180; with 4, the same as with 5, + 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: (n² + 3n + 2)(n² + 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) × (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 × 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.

[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.

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.

[Illustration:

1 2 +---------------+ +---------------+ \ R / \ B / \ B / \ R / \ / \ / \ / \ / \ / G \ / \ / G \ / \-------/ \-------/ \ / \ / \ Y / \ Y / \ / \ / ' '

3 4 5 +---------------+ +---------------+ +---------------+ \ R / \ R / \ R / \ G / \ Y / \ R / \ / \ / \ / \ / \ / \ / \ / G \ / \ / G \ / \ / G \ / \-------/ \-------/ \-------/ \ / \ / \ / \ Y / \ Y / \ Y / \ / \ / \ / ' ' '

6 7 8 +---------------+ +---------------+ +---------------+ \ G / \ Y / \ Y / \ Y / \ G / \ G / \ / \ / \ / \ / \ / \ / \ / G \ / \ / G \ / \ / G \ / \-------/ \-------/ \-------/ \ / \ / \ / \ Y / \ Y / \ Y / \ / \ / \ / ' ' '

]

282.--THE ANTIQUARY'S CHAIN.

[Illustration]

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.