Chapter XI: Conclusion (3)

+-----+-----+-----+-----+-----+
| | | | | |
| 1 | 8 | 6 | 7 | 9 |
|-----+-----+-----+-----+-----|
| | | | | |
| 2 | 1/6 | 1/2 | 1/4 | 1/8 |
|-----+-----+-----+-----+-----|
| | | | | |
| 3 | 2/4 | 1/8 | 2/1 | 2/7 |
|-----+-----+-----+-----+-----|
| | | | | |
| 4 | 3/2 | 2/4 | 2/8 | 3/6 |
|-----+-----+-----+-----+-----|
| | | | | |
| 5 | 4/0 | 3/0 | 3/5 | 4/5 |
|-----+-----+-----+-----+-----|
| | | | | |
| 6 | 4/8 | 3/6 | 4/2 | 5/4 |
|-----+-----+-----+-----+-----|
| | | | | |
| 7 | 5/6 | 4/2 | 4/9 | 6/3 |
|-----+-----+-----+-----+-----|
| | | | | |
| 8 | 6/4 | 4/8 | 5/6 | 7/2 |
|-----+-----+-----+-----+-----|
| | | | | |
| 9 | 7/2 | 5/4 | 6/3 | 8/1 |
+-----+-----+-----+-----+-----+
]

To use these in multiplication, select the strips, the top figures of which make the number to be multiplied. For example:

To multiply 8,679 by 8, look at the eighth line of squares from the top, and on that line will be found the product of each of the integers 8, 6, 7, 9, when multiplied by 8. We have then to write down the 2 as the first figure of the product, add 7 and 6 together = 13; write 3 as the next figure, carry 1 to add to the sum of 8 and 5, and so on.

The reason for dividing the figures in each square by a diagonal line, and for placing the left-hand figure higher than the right is, that the eye may be thus assisted in adding the carried figure of one slip to the unit of the next.

To provide for the occurrence of more than one of the same figures in the multiplicand, there should be several slips or rods for each of the digits.

In practice the rods are placed on a flat piece of wood, with two ridges at right angles, by which they are preserved in a proper position.

This instrument can be made useful in "divisions," by making by means of it a table of the product of the divisor, multiplied by each of the numbers 1 to 9.

THE ARITHMETICAL BOOMERANG.

The boomerang is an instrument of peculiar form, used by the natives of New South Wales, for the purpose of killing wild fowl and other small animals. If projected forwards, it at first proceeds in a straight line, but afterwards rises in the air, and after performing sundry peculiar gyrations, returns in the direction of the place where it was thrown.

The term is applied to those arithmetical processes by which you can divine a number thought of by another. You throw forwards the number by means of addition and multiplication, and then, by means of subtraction and division, you bring it back to the original starting point, making it proceed in a track so circuitous as to evade the superficial notice of the tyro.

TO FIND A NUMBER THOUGHT OF.

_First Method._

This is an arithmetical trick which, to those who are unacquainted with it, seems very surprising; but, when explained, it is very simple. For instance, ask a person to _think_ of any number under 10. When he says he has done so, desire him to treble that number. Then ask him whether the sum of the number he has thought of (now multiplied by 3) be odd or even; if odd, tell him to add 1 to make the sum even. He is next to halve the sum, and then treble that half. Again ask whether the amount be odd or even. If odd, add 1 (as before) to make it even, and then halve it. Now ask how many nines are contained in the remainder. The secret is, to bear in mind whether the first sum be odd or even; if odd, retain 1 in the memory; if odd a second time, retain 2 more (making in all 3 to be retained in the memory;) to which add 4 for every nine contained in the remainder.

For example, No. 7 is odd the first and also the second time; and the remainder (17) contains one nine; so that 1, added to 2, make 3, and 3, added to 4, make 7, the number thought of. No. 1 is odd the first time (retain 1), and even the second (of which no notice is taken), but the remainder is not equal to nine. No. 2 is even the first and odd the second time (retain 2), but the remainder contains no nine. No. 3 is odd the first and the second time, still there is no nine in the remainder. No. 4 is even both times, and contains one nine. No. 5 is odd the first time and the remainder contains one nine. No. 6 is odd the second time, and contains one nine in the remainder. No. 8 is even both times, and the remainder contains two nines. No notice need be taken of any overplus of a remainder, after being divided by nine.

The following are illustrations of the result with each number:

1 2 3 4 5 6 7 8 9
3 3 3 3 3 3 3 3 3
-- -- -- -- -- -- -- -- --
3 2)6 9 2)12 15 2)18 21 2)24 27
Add 1 -- Add 1 -- Add 1 -- Add 1 -- Add 1
-- 3 -- 6 -- 9 -- 12 --
2)4 3 2)10 3 2)16 3 2)22 3 2)28
-- -- -- -- -- -- -- -- --
2 9 5 2)18 8 27 11 2)36 14
3 Add 1 3 -- 3 Add 1 3 -- 3
-- -- -- 9)9 -- -- -- 9)18 --
2)6 2)10 15 -- 2)24 2)28 33 -- 2)42
-- -- Add 1 1 -- -- Add 1 2 --
3 5 -- 9)12 9)14 -- 9)21
2)16 -- -- 2)34 --
-- 1 1 -- 2
8 9)17
--
1

_Second Method._

EXAMPLE.

Let a person think of a number, say 6
1. Let him multiply it by 3 18
2. Add 1 19
3. Multiply by 3 57
4. Add to this the number thought of 63

Let him inform you what is the number produced; it will always end with 3. Strike off the 3, and inform him that he thought of 6.

_Third Method._

EXAMPLE.

Suppose the number thought of to be 6
1. Let him double it 12
2. Add 4 16
3. Multiply by 6 80
4. Add 12 92
5. Multiply by 10 920

Let him inform you what is the number produced. You must in every case subtract 320; the remainder is, in this example, 600; strike off the two ciphers, and announce 6 as the number thought of.

_Fourth Method._

Desire a person to think of a number, say 6. He must then proceed--

EXAMPLE.

1. To multiply this number by itself 36
2. So take 1 from the number thought of 5
3. To multiply this by itself 25
4. To tell you the difference between this
product and the former 11
You must then add 1 to it 12
And halve this number 6

Which will be the number thought of.

_Fifth Method._

Desire a person to think of a number, say 6. He must then proceed as follows:

EXAMPLE.

1. Add 1 to it 7
2. Multiply by 3 21
3. Add 1 again 22
4. Add the number thought of 28

Let him tell you the figures produced (28):

5. You then subtract 4 from it 24
6. And divide by 4 6

Which you can say is the number thought of.

_Sixth Method._

EXAMPLE.

Suppose the number thought of 6
1. Let him double it 12
2. Desire him to add to this any number
you tell him, say 4 16
3. To halve it 8

You can then tell him that if he will subtract from this the number he thought of, the remainder will be, in the case supposed, 2.

_Note._--The remainder is always half of the number you tell him to add.

TO DISCOVER TWO OR MORE NUMBERS THAT A PERSON HAS THOUGHT OF.

_1st Case._--Where each of the numbers is less than 10. Suppose the numbers thought of were 2, 3, 5.

EXAMPLE.

1. Desire him to double the first number making 4
2. To add 1 to it 5
3. To multiply by 5 25
4. To add the second number 28

There being a third number, repeat this process--

5. To double it 56
6. To add 1 to it 57
7. To multiply by 5 285
8. To add the third number 290

And to proceed in the same manner for as many numbers as were thought of. Let him tell you the last sum produced (in this case 290). Then, if there were two numbers thought of, you must subtract 5; if three, 55; if four, 555. You must here subtract 55, leaving a remainder of 235, which are the numbers thought of, 2, 3 and 5.

_2d Case._--Where one or more of the numbers are 10, or more than 10, and where there is an _odd_ number of numbers thought of.

Suppose he fixes upon five numbers, viz. 4, 6, 9, 15, 16.

He must add together the numbers as follows, and tell you the various sums:

1. The sum of the 1st and 2d 10
2. The sum of the 2d and 3d 15
3. The sum of the 3d and 4th 24
4. The sum of the 4th and 5th 31
5. The sum of the 1st and last 20

You must then add together the 1st, 3d and 5th sums, viz. 10 + 24 + 20 = 54, and the 2d and 4th, 15 + 31 = 46; take one from the other, leaving 8. The half of this is the 1st number, 4; if you take this from the sum of the 1st and 2d you will have the 2d number, 6; this taken from the sum of the 2d and 3d will give you the 3d, 9; and so on for the other numbers.

_3d Case._--Where one or more of the numbers are 10, or more than 10, and where an _even_ number of numbers has been thought of.

Suppose he fixes on six numbers, viz. 2, 6, 7, 15, 16, 18. He must add together the numbers as follows, and tell you the sum in each case:--

1. The sum of the 1st and 2d 8
2. The sum of the 2d and 3d 13
3. The sum of the 3d and 4th 22
4. The sum of the 4th and 5th 31
5. The sum of the 5th and 6th 34
6. The sum of the 2d and last 24

You must then add together the 2d, 4th and 6th sums, 13 + 31 + 24 = 68, and the 3d and 5th sums, 22 + 34 = 56. Subtract one from the other, leaving 12; the 2d number will be 6, the half of this; take the 2d from the sum of the 1st and 2d you will get the 1st; take the 2nd from the sum of the 2d and 3d, and you will have the 3d, and so on.

HOW MANY COUNTERS HAVE I IN MY HANDS?

A person having an equal number of counters in each hand, it is required to find how many he has altogether.

Suppose he has 16 counters, or 8 in each hand. Desire him to transfer from one hand to the other a certain number of them, and to tell you the number so transferred. Suppose it be 4, the hands now contain 4 and 12. Ask him how many times the smaller number is contained in the larger; in this case it is 3 times. You must then multiply the number transferred, 4, by the 3, making 12, and add the 4, making 16; then divide 16 by the 3 _minus_ 1; this will bring 8, the number in each hand.

In most cases fractions will occur in the process: when 10 counters are in each hand, and if 4 be transferred, the hands will contain 6 and 14.

He will divide 14 by 6 and inform you that the quotient is 2 3/6 or 2⅓.

You multiply 4 by 2⅓, which is 9⅓.

Add 4 to this, making 13⅓, equal to 40/3.

Subtract 1 from 2⅓, leaving 1⅓ or 4/3.

Divide 40/3 by 4/3 giving 10, the number in each hand.

THE MYSTERIOUS HALVINGS.

_To tell the number a person has thought of._

One of the company must fix upon any one of the numbers from 1 to 15; this he keeps secret, as well as the numbers produced by the succeeding operations:

Suppose he fixes on 8
He must add 1 to it, making 9
Triple it 27
Halve it[11]--1st _halving_--(_larger half_) 14
Triple it 42
Halve it--2d _halving_ 21
Triple it 63
Halve it--3d _halving_--(_larger half_) 32
Triple it 96
Halve it--4th _halving_. 48

He need not inform you that 48 is the figure produced, but he must let you know in which of the four halvings he was obliged to take a "larger half;" having ascertained this point, you discover the number fixed upon in the following manner. Carry in your mind, or on a slip of paper, the following list of names in which the letter a occurs in one or more of the three syllables of all except the last.

The three syllables are intended to represent the 1st, 2d, and 3d halvings, and the occurrence of the letter A corresponds to the occurrence of a "larger half" in one or more of these three halvings. Having been informed where the _larger half_ was taken, refer to the word which has A in the corresponding syllable, and against it stand two numbers, one of which was the number thought of; and of these two, the right hand number is the correct one _if a larger half was taken in the 4th stage_, and the left hand one _if the, 4th halving was exact_.

In the example given, a _larger half_ occurred in the 1st and 3d stage; this points us to _Car-ro-way_, and the halving in the 4th stage being exact, shows us that 8 was the number fixed upon.

If the 4th halving If a _larger half_ occurs
is _exact_. in the 4th halving.
WAsh-ing-ton 4 12
LA-fAy-ette 2 10
CAr-row-wAy 8 0
MAn-hAt-tAn 6 14
Ger-mA-ny 13 5
Tel-e-grAph 3 11
Bo-nA-pArte 1 9
Long-fel-low 15 7

It will be observed that there is always a difference of 8 between the numbers of the columns, so that it is necessary to recollect only one of them. Perhaps some of our readers who wish to be adepts in this game, would prefer recollecting the above table if put in this form:

2-3 1-2 3 1-2-3 1-3 2 none
--------------------------------
1 2 3 4 8 13 15

where the upper line denotes the cases in which the "larger half" was taken, and the lower line the numbers of the left hand column above given.

_Another Method._

The person having chosen any number from one to fifteen, he is to add twenty-one to that number, and triple the amount. Then,

1st. He is to take half of that triple, and triple that half.

2nd. To take the half of the last triple, and triple that half.

3rd. To take the half of the last triple.

4th. To take the half of the last half.

In this operation there are four distinct cases or stages where the half is to be taken. The three first are denoted by one of the eight following Latin words, each word being composed of three syllables, and the syllables containing the letter i corresponding in numerical order with the cases where the half cannot be taken without a fraction; consequently, in those cases the person who makes the deduction is to add one to the number to be divided. The fourth case shows which of the two numbers corresponding to each word has been chosen. For if the fourth half can be taken without adding one, the number chosen is in the first, or left-hand column; but if not, it is in the second column to the right.

The words. The numbers denoted.
Mi-ser-is 8 0
Ob-tin-git 1 9
Ni-mi-um 2 10
No-tar-i 3 11
In-fer-nos 4 12
Or-di-nes 13 5
Ti-mi-di 6 14
Te-ne-ant 15 7

_Example._--Suppose the number chosen to be nine, to which is to be added one, making ten, and which last, being tripled, gives thirty. Then:

1st case. The half of the triple is 15
which tripled, makes 45
2nd case. The half of that triple, 1
being added to make an
even number, is 23
and that tripled, makes 69
3rd case. The half of the last triple,
1 being added, is 35
4th case. The half of the last half, 1
being again added, is 18

Here we see, that in the second and third case, one had to be added, and, looking at the table, we find that the only corresponding word having an i in its second and third syllables is _Ob-tin-git_, which represents the figures one and nine. Then, as one had to be added in the fourth case, we know by the rule, that the figure in the second column, 9, is the one required. Observe, that if no addition be required at any of the four stages, the number thought of will be fifteen; and if one addition only be required at the fourth stage, the number will be seven.

WHO WEARS THE RING?

This is an elegant application of the principles involved in discovering a number fixed upon. The number of persons participating in the game should not exceed nine. One of them puts a ring on one of his fingers, and it is your object to discover--1st. The wearer of the ring. 2d. The hand. 3d. The finger. 4th. The joint.

The company being seated in order the persons must be numbered 1, 2, 3, &c.; the thumb must be termed the first finger, the fore finger being the second; the joint nearest the extremity must be called the first joint; the right hand is one, and the left hand two.

These preliminaries having been arranged, leave the room in order that the ring may be placed unobserved by you. We will suppose that the third person has the ring on the right hand, third finger, and first joint; your object is to discover the figures 3131.

Desire one of the company to perform secretly the following arithmetical operations:

1. Double the number of the person who has the
ring; in the case supposed, this will produce 6
2. Add 5 11
3. Multiply by 5 55
4. Add 10 65
5. Add the number denoting the hand 66
6. Multiply by 10 660
7. Add the number of the finger 663
8. Multiply by 10 6630
9. Add the number of the joint 6631
10. Add 35 6666

He must apprise you of the figures now produced, 6666; you will then in all cases subtract from it 3535; in the present instance there will remain 3131, denoting the person No. 3, the hand No. 1, the finger No. 3, and the joint No. 1.

PROBABILITIES.[12]

When we look around us at results happening daily, of the causes of which we are ignorant, we are led to regard them as isolated incidents subject to no law or rule; but could we see and understand the secret workings and connection existing between cause and effect, we might frequently discover that all works by rule. As it is, we may readily mark the boundaries, within which events must happen in very many instances; and do much to estimate their probability. We speak of _Chance_, as something without plan or design, but taking in a large range, our calculations will approximate closely to the truth. When we throw a copper into the air, the chances of "heads or tails," as the boys say, are equal, and though one or the other may occur most frequently for a few throws, in a large number, say a thousand, the results will be about equally divided. In this case the sides of the coin must be equal in weight, else it will be like the grumbler's bread and butter:

"I never had a piece of bread,
Particularly good and wide,
But fell upon the sanded floor,
And always on the buttered side."

Had he put on less butter, perhaps the sides would have been more equal in weight, and the probability of the buttered side being uppermost would have been increased. Disturbing causes unknown to us, may often shape the result; but in the absence of these, we may pretty accurately estimate our chances.

We see accidents from fire and flood, happening at times and points least expected; but the insurer has learned by observation to estimate probabilities, and by taking a wide range of country and a period of years, he does a comparatively safe business. Death takes the young and the old; but the life insurer has conned the bills of mortality and studied the ages of those who have died, until he can estimate at once the probability of duration of life, and determine what he can afford to pay for an annuity contingent on life, or engage for a present sum, or an annual sum paid for life, to pay the heirs at the death of the insured. In one instance his estimate may fall short, and in another exceed, but the average will be about right.

So, too, the man who deals in lotteries and games of chance, knows the data and calculates carefully the probabilities, and though "luck" may sometimes be against him, his estimates of probabilities are based on mathematical principles, and he is secure in being ultimately the gaining party.

How these chances are calculated, depends on the data in each case, and it is not within the range of our present plan to attempt more than giving a general idea of the subject; and this with any one of ordinary prudence, will be sufficient to prevent all intermeddling with lotteries and every other species of gambling. The probabilities are always against the casual operator, even if all be conducted fairly; what then must they be when fraud and dishonesty are superadded? It is downright swindling!

In lottery schemes generally, fifteen per cent. is reserved as profit, but this is a small part of what may be secured; yet even this amounts to a great deal. If a man were to draw a prize nominally of $100,000, fifteen thousand would be deducted at once, and he would be entitled to only $85,000. It is true that in his good fortune he would not probably regard the abatement, but that does not change the principle.

VARIATIONS.

It is obvious that if we have a number of single things arranged in any order, we may change the arrangement into a variety of forms, and in doing so, we may take all together, or we may take only part at once. For instance, we may arrange the six vowels, a e, i, o, u, y, in a great number of ways, as a e i o u y, a i e o u y, e a i o u y, &c., &c.; or we may form them into groups, as ae, io, uy, ai, eu, oy, &c.; or, we may take three, four, five, or, as above, all at a time; and it is reasonable to suppose that the number of possible changes may, in all cases, be calculated.

When all are taken together, the operation is called _Permutation_; but if a part only be taken, it is called either a _Variation_ or a _Combination_; a e, i o, u y, are distinct combinations, and are also considered one of the variations of two of which those six letters are susceptible; e a, o i, y u, are three other variations, but they are the same combinations; for a change of order will constitute a new variation but not a new combination; hence the number of variations will always exceed the number of combinations.

The doctrine of variations and combinations forms the basis of many forms of lotteries, and of other calculations used in practical life.

COMBINATIONS AND PERMUTATIONS.

"Combinations" are the different ways in which a certain number of things can be selected out of a larger number, when taken 1 at a time, 2 at a time, or any other number each time, but without regard to the order in which the selected numbers can be arranged among themselves. The latter is the province of "Permutation," which refers to the different ways in which a number can be selected out of one that is larger, and, _in addition to this_, to the different ways of _grouping_ these selected numbers.

Thus 4 things can be taken 2 at a time in 6 different ways; for instance, the letters a, b, c, d, can be taken 2 at a time thus, a and b, a and c, a and d, b and c, b and d, c and d; if we regard the _order_ of the selected letters we shall find that these 4 letters are capable of 12 different permutations, as ab, ba, ac, ca, ad, da, bc, cb, bd, db, cd, dc.

If we selected 3 letters at a time we could make 4 different selections, and 24 different changes of grouping.

The rule to compute the number of these different ways is very simple, but sometimes involves a multitude of figures.

To determine the number of permutations, commence with unity, and multiply by the successive terms of the natural series 1, 2, 3, &c., until the highest multiplier shall express the number of individual things. The last product will indicate the number of possible changes.

_Example 1._ How many changes can be made in the arrangement of 5 grains of corn, all of different colors, laid in a row?

_Solution._ 1 × 2 × 3 × 4 × 5 = 120, _Ans._

This may seem improbable, the number being so great, but if there were but a single grain more, the possible changes would be 720; and another would extend the limit to 5040; and so onward in a constantly increasing ratio. The reason, however, will be obvious on a little scrutiny. If there were but one thing, as _a_, it would admit of but one position; but if two, as _a b_, it would admit of two positions, _ab_, _ba_. If three things, as _a b c_, then they will admit of 1 × 2 × 3 = 6 changes, for the last two will admit of two variations, as _a b c_, _a c b_, and each of the three may successively be placed first, and two changes made to each of the others, so that 3 × 2 = 6, the number of possible changes. In the same way we may show that if there be four individual things, each one will be first in each of the six changes which the other three will undergo, and consequently, there will be 24 changes in all. In this way we might show that when there are 5 individual things, there will be 5 times as many changes as when there were but 4; and when 6, there will be 6 times as many changes as when there are only 5; and so on _ad infinitum_, according to the same law.

_Example 2._ In how many ways may a family of 10 persons seat themselves differently at dinner? _Ans._ 3,628,800.

When we consider that this would require a period of 9935-55/487 years, the mind is lost in astonishment. The story of the man who bought a horse at a farthing for the first nail in his shoe, a penny for the second, &c., is thrown into the shade; and we incline to doubt whether there is not some mistake; and yet on just such chances as one to all these, do gamblers constantly risk their money!

_Example 3._ I have written the letters contained in the word N I M R O D on 6 cards; being one letter on each, and having thrown them confusedly into a hat, I am offered $10 to draw the cards successively, so as to spell the name correctly. What is my chance of success worth? _Ans._ 1-7/18 cents.

_Example 4._ In order to form a lottery scheme, I have put into the wheel as many cards as I can put 4 letters of the word Charleston on, without having the same letters in the same order upon any two cards. I offer $100 to him who draws the card having on it the first four letters of the said word in their natural order (Char). What is the chance of drawing a prize worth?

There are 10 letters in the word, and the combination is of the 4th class; and, according to the mode of determining combinations with repetitions, we find the whole number of combinations of the 4th class which the word admits of is 210. Then he has one chance in 210 of drawing the letters Char, in _some_ order. The number of permutations of four individual things is 1 × 2 × 3 × 4 = 24, and 210 × 24 = 5040 his chance of drawing them in the right order, and $100 divided by 5040 gives _Ans._ 1 52/63 cents.

Suppose that the numbers from 1 to 78, inclusive, be placed upon 78 cards, and the cards placed in a wheel by which they are thoroughly mixed; and then 13 cards be successively drawn out, by a person who has no means of choosing, and the numbers on them registered. Suppose also that tickets have been issued, containing each three of the 78 numbers, but no two having _all_ the same numbers, and that he who holds the ticket having on it the first three drawn numbers in their regular order, shall be entitled to $100,000; what would the probability of drawing such a ticket be worth?

_Ans._ 21 5183/5858 cents.

_Note._--It is usual also, to give smaller prizes to the holders of tickets having the numbers in any order, or having any two or one of the drawn numbers. Lotteries may be arranged on a great diversity of plans, and in each the probability of drawing prizes will vary.

A speaks the truth 3 times in 4; B 4 times in 5, and C 6 times in 7. What is the probability of an event which A and B assert, and C denies? _Ans._ 140/143

Suppose a coin be thrown up, having two faces; what is the probability that the obverse (heads) side will fall upward, and what the reverse?

Here there are only two possible cases, and one favors each of the contingencies the probability of each will be 1/(1 + 1) = 1/2; there being no reason why one side should fall uppermost rather than the other.

What would be the probability of either side presenting upwards twice in two throws?

Here we have 4 possible cases, viz.:

Obverse and reverse;
Obverse both times;
Reverse and obverse;
Reverse both times.

Of the 4 possibilities there is only one which favors the turning up of the obverse twice in succession, and the same is true of the reverse, hence the probability of either is only 1/4.

In like manner we might show that the probability of the obverse presenting upwards three times in succession will be 1/8, or 1/2 × 1/2 × 1/2; the general principle being to multiply successively together the independent probabilities of an event for the fraction expressing the chance of all the events happening.

THE VISITORS TO THE CRYSTAL PALACE.

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.