PERMUTATION.

¶ XCII. PERMUTATION is the method of finding how many different ways any number of things may be changed.

1. How many changes may be made of the three first letters of the alphabet? In this example, had there been but two letters, they could only be changed twice; that is, a, b, and b, a; that is, 1 X 2 = 2; but, as there are three let ters, they may be changed 1 x 2 x 36 times, as follows:

Hence, To find the Number of different Changes or Permu tations, which may be made with any given Ňumber of different Things;

Multiply together all the terms of the natural series, from 1 up to the given number, and the last product will be the number of changes required.

2. How many different ways may the first five letters of the alphabet be arranged? A. 120.

3. How many changes may be rung on 15 bells? and in what time may they be rung, allowing 3 seconds to every round? A. 1307674368000 changes

3923023104000 seconds.

4. What time will it require for 10 boarders to seat themselves differently every day at dinner, allowing 365 days to the year? A. 9941335 years. 5. Of how many variations will the 26 letters of the alphabet admit? A. 403291461126605635584000008

POSITION

Is a rule which teaches, by the use of supposed numbers, to find true ones. It is divided into two parts, called Sing. and Double.

¶ XCIII.

SINGLE POSITION.

This rule teaches to resolve those questio whose results are proportional to their suppositions.

A schoolmaster, being asked how many scholars he had, replied, " If I nad as many more as I now have, one half as many more, one third, and one fourth as many more, I should have 296." How many had he?

Let us suppose he had 24
Then as many more
as many
as many

as many

We have now found that we did not suppose the right number. If we had, the amount would

24
have been 296. But 24 has been increased in the
12 | same manner to amount to 74, that some unknown
8 number, the true number of scholars, must be, to

amount to 296. Consequently, it is obvious, that 674 has the same ratio to 296 that 24 has to the true number. The question may, therefore, be solved 74 by the following statement:

As 74: 296: 24: 96, Ans.

This answer we prove to be right by increasing it by itself, one half itself, one third itself, and one fourth itself;

96

96

48

From these illustrations we derive the following

RULE.

Suppose any number you choose, and proceed with it in the same manner you would with the answer, to see if it were right.

Then say, As this result the result in the question :: the supposed number : number sought.

More Exercises for the Slate.

2. James lent William a sum of money on interest, and in 10 years it amounted to $1600; what was the sum lent? A. $1000.

3. Three merchants gained, by trading, $1920, of which A took a certain sum, B took three times as much as A, and C four times as much as B: what share of the gain had each?

A. A's share was $120; B's, $360; and C's, $1440.

4. A person having about him a certain number of crowns, said, if a third, a fourth, and a sixth of them were added together, the sum would be 45; how many crowns had he? A. 60.

5. What is the age of a person, who says, that if

of the years he has

lived be multiplied by 7, and of them be added to the product, the sum would be 292? A. 60 years.

6. What number is that, which, being multiplied by 7, and the product divided by 6, the quotient will be 14? A. 12.

23 *

DOUBLE POSITION.

1 XCIV. This rule teaches to solve questions by means of two supposed numbers.

In SINGLE POSITION, the number sought is always mu! uplied or divided by some proposed number, or increased or diminished by itself, or some known part of itself, a certain number of times. Consequently, the result will be proportional to its supposition, and but one supposition will be necessary; but, in DOUBLE POSITION, we employ two, for the results are not proportional to the suppositions.

1. A gentleman gave his three sons $10000, in the following manner: to the second $1000 more than to the first, and to the third as many as to the first and second. What was each son's part?

Let us suppose the share of the first, 10001
Then the second 2000

The shares of all the sons will, if oar supposition be correct, amount to $10000; but, as they amount to $6000 only, wo cali the error 4000.

The first error, then, is $4000, and the second $2000. Now, the difference between these errors would seem to have the same relation to the difference of the suppositions, as either of the errors would have to the difference betweer the supposition which produced it and the true number. We can easily make this statement, and ascertain whether it will produce such a result:

As the difference of errors, 2000 500, difference of suppositions :: either of the errors (say the first), 4000: 1000, the difference between its supposition and the true number. Adding this difference to 1000, the supposition, the amount is 2000 for the share of the first son then $3000 that of the second, $5000 that of the third, Ans. For 2000 + 3000 + 5000 = 10000, the whole

estate.

Had the supposition proved too great, instead of too small, it is manifest that we must have subtracted this difference.

The differences between the results and the result in the question are called errors: these are said to be alike, when both are either too great or too small; unlike, when one is too great, and the other too small.

From these illustrations we derive the following

RULE.

Suppose any two numbers, and proceed with each according to the manner described in the question, and see how much the result of each differs from that in the question

Then say, As the difference of the errors: the difference of the suppositions :: either error: difference between its supposition and the number sought.

More Exercises for the Slate.

2. Three persons disputing about their ages, says B, "I am 10 years older than A;" says C, "I am as old as you both :" now, what were their several ages, the sum of all of them being 100? Ans. A's, 20; B's, 30; C's, 50.

3. Two persons, A and B, have the same income; A saves of his yearly; but B, by spending $150 per annum more than A, at the end of 8 years, finds himself $400 in debt; what is their income, and what does each spend per annum?

First, suppose each had $200; secondly, $300; then the errors will be 400 and 200. A. Their income is $400; A spends 300, B $450.

4. There is a fish whose head is 8 feet long, his tail is as long as his head and half his body, and his body is as long as his head and tail; what is the whole length of the fish?

First, suppose his body 30; secondly, 28; the errors will then be 1 and 2.

A. 32 feet.

5. A laborer was hired 80 days upon this condition, that for every day he was idle he should forfeit 50 cents, and for every day he wrought he should receive 75 cents; at the expiration of the time, he received $25; now, how many days did he work, and how many days was he idle?

A. He worked 52 days, and was idle 28

MISCELLANEOUS EXAMPLES.

J. There is a room, one side of which is 20 feet long and 8 feet high; how many square feet are contained in that side?

This side is a regular parallelogram (T LXXIX.); and, to find the square contents, we have seen that we must multiply the length by the breadth; thus, 20 ft. x 8 ft. i60 sq. ft., Ans.

But, had we been required to find the square contents of half of this parallelogram, as uivided in the figure on the left, it is plain that, if we should multiply (20) the whole length by

of (8) the width, or, in this case, the height, the product would be the square contents in this half, that is, in the figure BCD; thus, of 8=4; then, 4 X 2080 sq. ft., which is precisely of 160, the square contents in the whole figure.

The half B C D is called a triangle, because it has, as you see, 3 sides and 3 angles, and because the line B C falls perpendicularly on CD; the angle at C is called a right angle; the whole angle, then, B CD, may properly be ced a right-angled triangle.

The line B C is called a perpendicular, CD the base, and D B the hypothenuse. Note.-Both the base and perpendicular are sometimes called the legs of the triangle.

The difference of the errors, when alike, will be one subtracted from the other; when unlike, one added to the other.

Hence, To find the Area of a Right-angled Triangle ;— Multiply the length of the base by the length of the perpendicular; the product will be the area required.

2. What is the area of a triangular piece of land, one side of which is 40 rods, and the distance from the corner opposite that side to that side 20 rods? Ans. × 40=400 rods.

Note.-To find the area of any irregular figure, divide it into triangles.

Hence, To find the Hypothenuse, when the Legs are given;Add the squares of the two legs together, and extract 'the square root of their sum.

When the Hypothenuse and One Leg are given, to find the Other Leg;

From the square of the hypothenuse subtract the square of the given leg, and the square root of the remainder will be the other.

3. A river, 80 yards wide, passes by a fort, the walls of which are 60 yards high; now, what is the distance from the top of the wall to the opposite bank of the river?

In this example, we are to find the hypothenuse. Ans. 100 yards.

4. There is a certain street, in the middle of which, if a ladder 40 feet long be placed, it will reach a window 24 feet from the ground, on either side of said street; what is the width of the street?

In this example, we are to find the length of the base of two triangles, and then the sum of these will be the distance required. Ans. 64 feet.

5. There is a certain elm, 20 feet in diameter, growing in the centre of a circular island; the distance from the top of the tree to the water, in a straight line, is 120 feet; and the distance from the foot 90 feet; what is the height of the tree?

As the tree is 20 feet in diameter, the distance from its centre to the water is the length of the base, that 18, 10+ 90 100 feet. A. 66,332 ft.+.

6. Two ships sail from the same port; one goes due north 40 leagues, the other due east 30 leagues; how far are they apart?

We are here to find the hypothenuse. A. 50 leagues.