...

(A B) (C D) (E F) (G H) (I J) (K L).

Then give any pairing you like for the next day, say--

(A C) (B D) (E G) (F H) (I K) (J L),

and so on, until you have completed your eleven lines, with no pair ever occurring twice. There are a good many different arrangements possible. Try to find one of them. (2/2)

(1/2)

SOLUTION TO 32. THE EXCURSION TICKET PUZZLE. (5/5)

...

Florin in 3,818 ways.

Half-crown in 8,709 ways.

Double florin in 60,239 ways.

Crown in 166,651 ways.

Half-sovereign in 6,261,622 ways.

Sovereign in 500,291,833 ways.

It is a little surprising to find that a sovereign may be changed in over five hundred million different ways. But I have no doubt as to the correctness of my figures.

SOLUTION TO 32. THE EXCURSION TICKET PUZZLE. (4/5)

...

the other coins.

I will, however, add an interesting little table of the possible ways of changing our current coins which I believe has never been given in a book before. Change may be given for a

Farthing in 0 way.

Halfpenny in 1 way.

Penny in 3 ways.

Threepenny-piece in 16 ways.

Sixpence in 66 ways.

Shilling in 402 ways.

SOLUTION TO 32. THE EXCURSION TICKET PUZZLE. (3/5)

...

--------------------- + 1 ways.

18

If sixpences are also used there are

n^{4} + 22n³ + 159n² + 414n + 216

---------------------------------

216

ways, when the sum is a multiple of sixpence, and the constant, 216,

changes to 324 when the money is not such a multiple. And so the formulas increase in complexity in an accelerating ratio as we go on to

SOLUTION TO 32. THE EXCURSION TICKET PUZZLE. (2/5)

...

consider it worth the labour of working out.

Just to give an idea of what such a solution would involve, I will merely say that I find that, dealing only with those sums of money that are multiples of threepence, if we only use bronze coins any sum can be paid in (n + 1)² ways where n always represents the number of pence. If threepenny-pieces are admitted, there are

2n³ + 15n² + 33n

SOLUTION TO 32. THE EXCURSION TICKET PUZZLE. (1/5)

Nineteen shillings and ninepence may be paid in 458,908,622 different ways.

I do not propose to give my method of solution. Any such explanation would occupy an amount of space out of proportion to its interest or value. If I could give within reasonable limits a general solution for all money payments, I would strain a point to find room; but such a solution would be extremely complex and cumbersome, and I do not

...

"is nineteen shillings and ninepence, and I should like to know in just how many different ways it is possible for such an amount to be paid in the current coin of this realm."

Here, then, is a puzzle: In how many different ways may nineteen shillings and ninepence be paid in our current coin? Remember that the fourpenny-piece is not now current. (3/3)

...

Metropolis. The booking clerk was unaccustomed to dealing with crowds of such a dimension, and he told me afterwards, while wiping his manly brow, that what caused him so much trouble was the fact that these rustics paid their fares in such a lot of small money.

He said that he had enough farthings to supply a West End draper with change for a week, and a sufficient number of threepenny pieces for the congregations of three parish churches. "That excursion fare," said he, (2/3)

SOLUTION TO 269. THREE MEN IN A BOAT. (7/7)

...

and showed what that general law is and how the groups should be posed for any number of girls. I gave actual arrangements for numbers that had previously baffled all attempts to manipulate, and the problem may now be considered generally solved. Readers will find an excellent full account of the puzzle in W.W. Rouse Ball's _Mathematical Recreations_,

5th edition.

SOLUTION TO 269. THREE MEN IN A BOAT. (6/7)

...

solution of this puzzle had exercised the ingenuity of mathematicians since 1850, when the question was first propounded, until recently. In 1908 and the two following years I indicated (see _Educational Times Reprints_, Vols. XIV., XV., and XVII.) that all our trouble had arisen from a failure to discover that 15 is a special case (too small to enter into the general law for all higher numbers of girls of the form 6n+3),

SOLUTION TO 269. THREE MEN IN A BOAT. (5/7)

...

It will be found that no two men ever go out twice together, and that no man ever goes out twice in the same boat.

This is an extension of the well-known problem of the "Fifteen Schoolgirls," by Kirkman. The original conditions were simply that fifteen girls walked out on seven days in triplets without any girl ever walking twice in a triplet with another girl. Attempts at a general

SOLUTION TO 269. THREE MEN IN A BOAT. (4/7)

...

3 5 4 1 2

3rd Day (AJM) (BEH) (CFI) (DKO) (GNL)

7 6 8 9 1

4th Day (AEK) (CGM) (BOI) (DHL) (JNF)

4 5 3 10 2

5th Day (AHN) (CDJ) (BFL) (GEO) (MKI)

6 7 8 10 1

6th Day (AFO) (BGJ) (CKH) (DNI) (MEL)

5 4 3 9 2

7th Day (AIL) (BDM) (CEN) (GKF) (JHO)

SOLUTION TO 269. THREE MEN IN A BOAT. (3/7)

...

Of course only a certain number of all these arrangements are available when we have that other condition of using the smallest possible number of boats. As a matter of fact we need employ only ten different boats. Here is one the arrangements:--

1 2 3 4 5

1st Day (ABC) (DBF) (GHI) (JKL) (MNO)

8 6 7 9 10

2nd Day (ADG) (BKN) (COL) (JEI) (MHF)

SOLUTION TO 269. THREE MEN IN A BOAT. (2/7)

...

the reader will realize why this must be, for although, as an example, A must go out once with B and once with C, it does not necessarily follow that he must go out with C on the same occasion that he goes with B. He might take any other letter with him on that occasion, though the fact of his taking other than B would have its effect on the arrangement of the other triplets.

SOLUTION TO 269. THREE MEN IN A BOAT. (1/7)

If there were no conditions whatever, except that the men were all to go out together, in threes, they could row in an immense number of different ways. If the reader wishes to know how many, the number is 455^7. And with the condition that no two may ever be together more than once, there are no fewer than 15,567,552,000 different solutions--that is, different ways of arranging the men. With one solution before him,

...

The men within each pair of brackets are here seen to be in the same boat, and therefore A can never go out with B or with C again, and C can never go out again with B. The same applies to the other four boats. The figures show the number on the boat, so that A, B, or C, for example,

can never go out in boat No. 1 again. (5/5)

...

readers to discover how the men should have been placed in the boats. As their names happen to have been Andrews, Baker, Carter, Danby, Edwards,

Frith, Gay, Hart, Isaacs, Jackson, Kent, Lang, Mason, Napper, and Onslow, we can call them by their initials and write out the five groups for each of the seven days in the following simple way:

1 2 3 4 5

First Day: (ABC) (DEF) (GHI) (JKL) (MNO).

(4/5)

...

twice in the same boat. If you can manage to do this, and use as few different boats as possible, you may charge the firm with the expense."

One of the men tells me that the experience he has gained in such matters soon enabled him to work out the answer to the entire satisfaction of themselves and their employer. But the amusing part of the thing is that they never really solved the little mystery. I find their method to have been quite incorrect, and I think it will amuse my (3/5)

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Puzzles from Henry Ernest Dudeney's "Amusements in Mathematics"

Source: https://www.gutenberg.org/ebooks/16713

Joined Sep 2017