Part 21
In a family consisting of 8 young people, it was agreed that 3 at a time should visit the Crystal Palace, and that the visit should be repeated each day as long as a different trio could be selected. In how many days were the possible combinations of 3 out of 8 completed?
We must multiply 8 × 7 × 6, and also 3 × 2 × 1, and divide the product of the former, 336, by the product of the latter, 6; the result is 56, the number of visits, a different three going each time. So much gratified were they with the results of their agreement, that they wished to be allowed another series of visits, to be continued as many days as they could group 3 together in different order when starting. If Paterfamilias had granted such permission he would have had to wait 56 multiplied by 3 × 2 × 1, or 336 days, before this "new series" of visits would have come to a _finis_.
HOW MANY CHANGES CAN BE GIVEN TO 7 NOTES OF A PIANO?
That is to say, in how many ways can 7 keys be struck in succession, so that there shall be some difference in the order of the notes each time?
The result of multiplying
7 × 6 × 5 × 4 × 3 × 2 × 1
is 5,040, the number of changes.
THE ARITHMETICAL TRIANGLE.
This name has been given to a contrivance said to have originated with the famous Pascal, or to have been perfected by him.
1 2 1 3 3 1 4 6 4 1 5 10 10 5 1 6 15 20 15 6 1 7 21 35 35 21 7 1 8 28 56 70 56 28 8 1 &c. &c.
This peculiar series of numbers is thus formed: Write down the numbers 1, 2, 3, &c., as far as you please, in a vertical row. On the right hand of 2 place 1, add them together, and place 3 under the 1; then 3 added to 3 = 6, which place under the 3; 4 and 6 are 10, which place under the 6, and so on as far as you wish. This is the second vertical row, and the third is formed from the second in a similar way.
This triangle has the property of informing us, without the trouble of calculation, how many combinations can be made, taking any number at a time out of a larger number.
Suppose the question were that just given; how many selections can be made of 3 at a time out of 8? On the horizontal row commencing with 8, look for the third number; this is 56, which is the answer.
HOW MANY DIFFERENT DEALS CAN BE MADE WITH 13 CARDS OUT OF 52?
To discover this we must make a continued multiplication of 52 × 51 × 50 × 49 × 48 × 47 × 46 × 45 × 44 × 43 × 42 × 41 × 40, being 13 terms for the 13 cards, also a continued multiplication of 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, and having found the two products, we must divide one by the other, and the quotient is the number of different deals out of 52 cards. This "sum," that looks so formidable with natural figures, is a very short one by logarithms.
THE THREE GRACES.
Three articles, or three names inscribed on cards, having been distributed between three persons, you are to tell which article or card each person has.
Designate the three persons in your own mind, as 1st, 2d, and 3d, and the three articles, A, E, I. Provide 24 counters, and give 1 to the first person, 2 to the 2d, 3 to the 3d. Place the remaining 18 on the table. Request that the three persons will distribute among themselves the three articles, and that, having done so, the person who has the one which you have secretly denoted by A, will take as many counters as he may have already; the holder of E must take twice as many as he may have; and the holder of I must take four times as many. Then leave the room, in order that the distribution of articles and of counters may be made unobserved by you. We will suppose that the three articles are three cards, on which are the words Clara, Rosa, Emily, which you will yourself secretly denote by the letters A, E, I. Suppose also that in the division the first person has Emily (I), the second has Clara (A), and the third has Rosa (E), then the 1st will take four times as many counters as he has (1), and will therefore take 4; the 2d will take as many as he has (2), and will therefore take 2; the 3d will take 6, being twice as many as he has (3). On the table will be left 6 counters. The distribution having been made, you will return and observe the number of counters on the table, from which you can find who is the holder of each card by the following method.
It is plain that if the cards held by the 1st and 2d can be told, that held by the 3d will be known. It will be found that only six numbers can remain, viz. 1, 2, 3, 5, 6, 7; never 4, and never more than 7. Now the 6 combinations of a, e, and i, here given, represent the articles held by the 1st and 2d persons.
1 2 3 4 5 6 7 ae ea ai -- ei ia ie
In the case supposed, 6 counters being on the table, the combination _ia_ indicates that the first person has the card you have called I (Emily), the 2d has A (Clara), so that the 3d has E (Rosa).
In order to recollect the combinations of A, E, and I, it will be best to keep in memory some 7 words which form a sentence, and which contain these vowels in the order just given.
Our young friends can amuse themselves in forming a sentence for themselves, but as examples we supply three.
1 2 3 4 5 6 7 _ae_ _ea_ _ai_ -- _ei_ _ia_ _ie_ James easy admires now reigning with a bride. Anger, fear, pain may be hid with a smile. Graceful Emma, charming she reigns in all circles.
Or, if they prefer Latin, they can use the pentameter made up by the inventor of this beautiful pastime:
1 2 3 5 6 7 Salve certa animæ semita vita quies.
ANOTHER METHOD.
The performer must mentally distinguish the articles by the letters A, B, C, and the persons as 1st, 2d, and 3d. The persons having made their choice, give 12 counters to the 1st, 24 to the 2d, and 36 to the 3d. Then request the 1st person to add together the half of the counters of the person who has chosen A, the 3d of the person who has chosen B, and the 4th of those of the person who has chosen C, and then ask the sum, which must be either 23, 24, 25, 27, 28, or 29, as in the following table:
First. Second. Third. 12 24 36 A B C 23 A C B 24 B A C 25 C A B 27 B C A 28 C B A 29
This table shows that if the sum be 25, for example, the 1st person must have chosen B, the 2d A, and the 3d C; or if it be 28, the 1st must have chosen B, the 2d C, and the 3d A.
ANOTHER METHOD.
Three things having been divided between three persons, you are to determine the holder of each.
Call the persons in your own mind 1st, 2d, 3d.
Give to the 1st a card on which you have written the number 12; to the 2d the number 24; to the 3d 36.
The three things you must denote as A, E, I.
To simplify it you may have three cards with a name upon each, of which the initial letters are A, E, I, as Anna, Emma, Isabel.
Request your friends to divide between them the three articles, and then to add together certain parts of the numbers on their cards, as follows:
Whoever has A must supply one half of the number on his card;
Whoever has E must supply one third;
Whoever has I must supply one fourth;
This half, third and fourth having been added together, the sum must be announced to you on your return; and from this number you can tell who has A, who has E, and who has I.
If the No. is the 1st has the 2d has the 3d has 23 A E I 24 A I E 25 E A I 27 I A E 28 E I A 29 I E A
The sum which will be given to you can be one of six only. There are only six ways in which the articles can be divided, and there is a definite number for each of them.
The number 26 can never occur, and to recollect the six which do occur, and which you perceive are consecutive, you need take note only of what the 1st and 2d persons have.
23 24 25 26 27 28 29 ae ai ea -- ia ei ie
If you make up a line of good (or bad) English, having the vowels in the order here given, you will find it will aid you in their recollection. We give one as a specimen:
ae ai ea -- ia ei ie Brave dashing sea, like a giant revives itself.
THE FORTUNATE NINTH.
A sharp youth, fresh from school, having gone to visit a good-natured uncle, the latter placed on a table fifteen fine oranges and fifteen apples, and desired his young friend to take half. He, not liking the apples, was about to take the fifteen oranges; but this monopoly of the best fruit being objected to, the old gentleman told him to range all the fruit in a circle, and to take every ninth. The clever fellow ranged them in such a way as that, by taking away every ninth, all the apples were left on the table, and all the oranges were transferred to his capacious pockets. How did he arrange them?
He placed them as in the margin, A representing apples, and O oranges; and it will be found that, by commencing at the four apples, and going round and round the circle, taking away every ninth, all the oranges will be removed, and all the apples will remain.
If we let the vowels a e i o u denote the figures 1 2 3 4 5,
the arrangement of the figures 4, 5, 2, 1, &c., can be easily recollected by the following line:
Our earth's final fate--enigma ever dark 45 21 3 1 1 2 2 3 1 2 2 1 or, Our dear Richard's tale begins at the sea. 45 21 3 1 1 2 2 3 1 2 21
Our young friends may find amusement in forming lines for themselves as much superior to these as possible.
THE TEN TENS.
Take ten pieces of card, and upon each write any ten words; there is no restriction as to the initial letter of nine of the words, but the last word on each card must commence with certain letters which you must in your own mind associate with the numbers 1 to 10, so that by knowing the initial letter of the last word on each card, you can determine its number.
Here are ten cards, (call these the _Selecting Cards_,) which we give by way of example, though our readers will perhaps prefer having words of their own selection.
Jane. Ellen. George. James. Newton. Mary. Fanny. William. Clement. Davy. Matilda. Caroline. Frederick. Edward. Morse. Sarah. Isabel. Robert. Ralph. Fulton. Rosa. Flora. Edmund. Francis. Franklin. Elizabeth. Laura. John. Edwin. Arago. Harriet. Maria. Alfred. Walter. Spurzheim. Ann. Frances. Albert. Charles. Laplace. Emily. Edith. Henry. Samuel. Steers. =E=mma. =D=orothea. =I=saac. =T=heodore. =H=erschel.
Sister. Rose. Friendship. Putnam. Clay. Brother. Violet. Happiness. Lafayette. Webster. Uncle. Lupin. Industry. Steuben. Calhoun. Aunt. Daisy. Ambition. Scott. Benton. Grandmother. Tulip. Energy. Taylor. Jefferson. Grandfather. Peony. Fidelity. Green. Adams. Nephew. Hyacinth. Affection. Harrison. Madison. Niece. Pink. Hope. Hamilton. Jackson. Cousin. Snowdrop. Justice. Wayne. Monroe. =F=ather. =L=ily. =O=rder. =W=ashington. =N=apoleon.
For these the key words are, "Edith Flown," so that the letters
E D I T H F L O W N Stand for 1 2 3 4 5 6 7 8 9 10
For the success of the game, the key words and the numbers denoted by their letters, must be carefully concealed.
Take ten other cards, which call the "_grouped cards_," and upon one write down the first word from each of the selecting cards, being careful to write them in the same order. Let another card contain all the words which are second from the top, and so on till all the words have been grouped together. As an example, we give the 1st and 4th grouped cards.
1st. 4th. Jane. Sarah. Ellen. Isabel. George. Robert. James. Ralph. Newton. Fulton. Sister. Aunt. Rose. Daisy. Friendship. Ambition. Putnam. Scott. Clay. Benton.
The object of the game is to guess which of the words from any of the _selecting cards_ any person may have fixed upon.
Let any one choose a card out of the _selecting cards_, and after he has fixed upon a word, give it back to you; when receiving it, carefully note the last word upon it, which will give you, by the aid of the key word, the number of the card; this you must keep secret, and you then give him all the _grouped cards_, and request him to show you the cards which contain the words he fixed upon.
You can then announce the word; for the number of the word from the top on the grouped card is the same as the number of the selecting card, from which he made his choice.
Suppose he made his choice from the card which has Theodore for its last word--this is No. 4; when he shows you the grouped card, which he says contains the selected word, you will know that Ralph, the fourth from the top, is the name he fixed upon.
DIVIDING THE BEER.
During the siege of Sebastopol, when the troops were on "short allowance," a can of eight pints of porter was ordered to be equally divided between two messes; but having only a five pint can, and one which held three pints, it was found impossible to make this division, till one of the clever sappers suggested the following method; and, to understand it, we will put down the contents of each of the three cans at each stage of the process; commencing with
8-pt. 5-pt. 3-pt. The 8-pint can full, and the others empty, 8 0 0 1. Filled the 5-pint can 3 5 0 2. Filled the 3-pint can from the 5-pint 3 2 3 3. Pour the contents of the 3-pint into the 8-pint 6 2 0 4. Transfer the 2 pints from the 5-pint to the 3-pint 6 0 2 5. Filled the 5-pint from the 8-pint 1 5 2 6. Fill up the 3-pint from the 5-pint 1 4 3 7. Poured the 3 pints into the 8-pint; completing the feat 4 4 0
This was a dexterous expedient of the worthy sapper, the only objections to it being the time the thirty men had to wait, and the resulting flat condition of the beer.
THE DIFFICULT CASE OF WINE.
A gentleman had a bottle containing 12 pints of wine, 6 of which he was desirous of giving to a friend; but he had nothing to measure it, except two other bottles, one of 7 pints, and the other of 5. How did he contrive to put 6 pints into the 7-pint bottle?
12-pt. 7-pt. 6-pt. Before he commenced, the contents of the bottles were 12 0 0 1. He filled the 5-pint 7 0 5 2. Emptied the 5-pint into the 7-pint 7 5 0 3. Filled again the 5-pint from the 12-pint 2 5 5 4. Filled up the 7-pint from the 5 2 7 3 5. Emptied the 7-pint into the 12-pint 9 0 3 6. Poured the 3 pints from the 5 into the 7 9 3 0 7. Filled the 5-pint from the 12-pint 4 3 5 8. Filled up the 7-pint from the 5-pint 4 7 1 9. Emptied the 7-pint into the 12-pint 11 0 1 10. Poured 1 pint from the 5-pint into the 7-pint 11 1 0 11. Filled the 5-pint from the 12-pint 6 1 5 12. Poured the contents of the 5-pint into the 7-pint 6 6 0
ANOTHER DECIMATION OF FRUIT.
On the next visit of the youth to his uncle, the latter produced thirty apples and ten oranges, and offered him the favorite oranges, if his nephew could arrange them in an oval, so that by taking every twelfth the apples should remain. But this he could not accomplish, and the old gentleman, being well versed in the "Recreations in Science," proceeded to arrange them thus:
The places which the oranges here occupy can be easily remembered, being Nos. 7, 8, 11, 12, 21, 22, 24, 34, 36, 37.
THE WINE AND THE TABLES.
A certain hotel-keeper was dexterous in contrivances to produce a large appearance with small means. In the dining-room were three tables, between which he could divide 21 bottles, of which 7 only were full, 7 half full, and 7 apparently just emptied, and in such a manner that each table had the same number of bottles, and the same quantity of wine. He did this in two ways:
Table. Full Hf. full. Empty. Table. Full Hf. full. Empty. 1 2 3 2 | 1 3 1 3 2 2 3 2 | 2 3 1 3 3 3 1 3 | 3 1 5 1
He also performed a similar exploit with 24 bottles, 8 full, 8 half-full, and 8 empty:
Table. Full Hf. full. Empty. Table. Full Hf. full. Empty. 1 3 2 3 | 1 2 4 2 2 3 2 2 | 2 2 4 2 3 2 4 2 | 3 4 0 4
Also with 27 bottles, 9 full, 9 half-full, and 9 empty:
Table. Full Hf. full. Empty. Table. Full Hf. full. Empty. 1 2 5 2 | 1 1 7 1 } 2 3 3 3 | 2 4 1 4 } 3 4 1 4 | 3 4 1 4 }
THE THREE TRAVELERS.
Three men met at a caravansary or inn, in Persia; and two of them brought their provision along with them, according to the custom of the country; but the third not having provided any, proposed to the others that they should eat together, and he would pay the value of his proportion. This being agreed to, A produced 5 loaves, and B 3 loaves, all of which the travelers ate together, and C paid 8 pieces of money as the value of his share, with which the others were satisfied, but quarreled about the division of it. Upon this the matter was referred to the judge, who decided impartially. What was his decision?
At first sight it would seem that the money should be divided according to the bread furnished; but we must consider that, as the 3 ate 8 loaves, each one ate 2⅔ loaves of the bread he furnished. This from 5 would leave 2⅓ loaves furnished the stranger by A; and 3 - 2⅔ = ⅓ furnished by B, hence 2⅓ to ⅓ = 7 to 1, is the ratio in which the money is to be divided. If you imagine A and B to furnish, and C to consume all, then the division will be according to amounts furnished.
WHICH COUNTER HAS BEEN THOUGHT OF OUT OF SIXTEEN?
Take sixteen pieces of card, and number them 1 to 16. Arrange them in two rows, as at A B.
A B C B D M E B F N G B H 1 9 1 9 2 2 2 9 4 2 2 9 6 2 10 3 10 4 4 6 10 8 6 1 10 5 3 11 5 11 6 6 1 11 3 1 4 11 8 4 12 7 12 8 8 5 12 7 5 3 12 7 5 13 13 1 13 4 13 6 14 14 3 14 8 14 7 15 15 5 15 3 15 8 16 16 7 16 7 16
Desire a person to think of one of the numbers, and to tell you in which row it is. Suppose he fixes on 6; he will tell you that the row A contains the number he thought of.
Take up the row A, and arrange the numbers on each side of the row B, as shown at C D, so that the first number of the row A may be the first of the row C, the second of A be the first of D, the third of A be the second of C, and so on.
Ask in which of the rows, C or D, is the number thought of: in the case supposed it is in D.
Take up the rows C D, and put one underneath the other, as at M, taking care that the half-row in which is the number thought of, shall be above the other.
Divide it again into two rows, as at E F, on each side of B, in the same way as before. Ask again in which row it is: it is now in E.
Place one row under the other, as at N, and divide again into two rows, which will now be as G H.
You will be informed that the number is in row H, and you may then announce it to be the top number of that row.
The number thought of will always be _at the top of one of the rows after three transpositions_. If there were 32 counters it would be at the top after four transpositions.
MAGIC SQUARES.
The name "Magic Square" is given to a square divided into several smaller squares, in which numbers are placed in such a manner that every column of numbers, whether vertical, horizontal, or from corner to corner, shall amount to the same sum.
They are divided into three principal classes: 1st, Those which have an odd number of squares in each band; 2d, Those which have an even number of squares in each band, this even number being divisible exactly by 4; 3d, Where the even number of squares in each band cannot be divided by 4 without a fraction.
ODD MAGIC SQUARES.
Squares of this kind are formed thus. Imagine an exterior line of squares above the magic square you wish to form, and another exterior line on the right hand of it These two imaginary lines are shown in the figure.
Then attend to the two following rules:
1st. In placing the numbers in the squares we must go in an ascending oblique direction from left to right; any number which, by pursuing this direction, would fall into the exterior line, must be carried along that line of squares, whether vertical or horizontal, to the last square. Thus, 1 having been placed in the center of the top line, (see the first table on p. 228,) 2 would fall into the exterior square above the fourth vertical line; it must be therefore carried down to the lowest square of that line; then, ascending obliquely 3 falls into the square, but four falls out of it, to the end of a horizontal line, and it must be carried along that line to the extreme left, and there placed. Resuming our oblique ascension to the right, we place 5, where the reader sees it, and would place 6 in the middle of the top band, but finding it occupied by 1, we look for direction to the 2d Rule, which prescribes that, when in ascending obliquely, we come to a square already occupied, we must place the number, which according to the first rule should go into that occupied square, directly under the last number placed. Thus, in ascending with 4, 5, 6, the 6 must be placed directly under the 5, because the square next to 5 in an oblique direction is "engaged."