Then multiply each of the said errors by the contrary supposition, namely, the first position by the second error, and the second position by the first error. Then,

If the errors are alike, divide the difference of the products by the difference of the errors, and the quotient will be the

answer.

But if the errors are unlike, divide the sum of the products by the sum of the errors, for the answers.

Note, The errors are said to be alike, when they are either both too great or both too little ; and unlike, when one is too great and the other too little.

EXAMPLES.

1. What number is that, which being multiplied by 6, the product increased by 18, and the sum divided by 9, the quotient shall be 20?

Suppose the two numbers 18 and 30. Then,

First Position.

Second Position.

Proof.

27

6 mult.

FIND, by trial, two numbers, as near the true number as convenient, and work with them as in the question; marking the errors which arise from each of them.

Multiply the difference of the two numbers assumed, or found by trial, by one of the errors, and divide the product by the difference of the errors, when they are alike, but by their sum when they are unlike.

Add

Add the quotient, last found, to the number belonging to the said error, when that number is too little, but subtract it when too great, and the result will give the true quantity Sought*.

EXAMPLES.

1. So, the foregoing example, worked by this 2d rule will be as follows:

30 positions 18;

-2 errors + 6;

their dif. 12

least error 2

sum of errors 8) 24 (3 subtr.
from the position 30

leaves the answer 27

Ex. 2. A son asking his father how old he was, received this answer: Your age is now one-third of mine; but 5 years ago, your age was only one-fourth of mine. What then are their two ages? Ans. 15 and 45.

3. A workman was hired for 20 days, at 38 per day, for every day he worked; but with this condition, that for every day he played, he should forfeit 18. Now it so happened, that upon the whole he had 2/ 4s to receive. How many of the days did he work? Ans. 16.

4. A and B began to play together with equal sums of money: A first won 20 guineas, but afterwards lost back of what he then had; after which, в had 4 times as much as What sum did each begin with? Ans. 100 guineas.

A.

5. Two persons, A and B, have both the same income, a saves of his; but в, by spending 50l per annum more than

A, at the end of 4 years finds himself 1007 in debt.

What does each receive and spend per annum ?

Ans. They receive 1251 per annum; also a spends 1007, and B spends 150l per annum.

*For since, by the supposition, r: s:: x-a: x-b, therefore by division, gings: ba: ab, which is the 2d Rule.

PERMUTATIONS

PERMUTATIONS AND COMBINATIONS.

PERMUTATION is the altering, changing or varying the position or order of things; or the showing how many different ways they may be placed.-This is otherwise called Alternation, Changes, or Variation; and the only thing to be regarded here, is the order they stand in; for no two parcels are to have all their quantities placed in the same situation; as, how many changes may be rung on a number of bells, or how many different ways any number of persons may be placed, or how many several variations may be made of any number of letters, or any other things proposed to be varied.

COMBINATION is the showing how often a less number of things can be taken out of a greater, and combined together, without considering their places, or the order they stand in. This is sometimes called Election or Choice; and here every parcel must be different from all the rest, and no two are to have precisely the same quantities or things.

Combinations of the same Form, are those in which there are the same number of quantities, and the same repetitions : thus, aabc, bbcd, ccde, are of the same form; aabc, abbb, aabb, are of different forms.

Composition of Quantities, is the taking a given number of quantities out of as many equal rows of different quantities, one out of every row, and combining them together.

Illustrations of these definitions are in the following Problems:

PROBLEM I.

To assign the Number of Permutations, or Changes, that can be made of any Given Number of Things, all different from each other.

RULE*.

MULTIPLY all the terms of the natural series of numbers, from 1 up to the given number, continually together, and the last product will be the answer required.

EXAMPLES.

*The reason of the Rule may be shown thus; any one thing a is capable only of one position, as a.

Any two things a and b, are only capable of two variations; as ab, ba; whose number is expressed by 1×2.

EXAMPLES.

1. How many changes may be rung on 6 bells.

1

2

23

6

4

24

5

120

6

720 the Answer.

Or 1X2X3X4X5×6=720 the Answer.

2. How many days can 7 persons be placed in a different position at dinner? Ans. 5040 days.

3. How many changes may be rung on 12 bells, and what time would it require, supposing 10 changes to be rung in 1 minute, and the year to consist of 365 days, 5 hours, and 49 minutes ?

Ans. 479001600 changes, and 91 years, 26 days, 22 hours, 41 minutes.

4. How many changes may be made of the words in the following verse: Tot tibi sunt dotes, virgo, quot sidera cœlo ? Ans. 40320 changes.

If there be three things, a, b. and c; then any two of them, leaving out the 3d, will have 1X2 variations; and consequently when the 3d is taken in, there will be 1X2X3 variations.

In the same manner, when there are 4 things, every three, leaving out the 4th, will have 1×2×3 variations; consequently by taking in successively the 4 left out, there will be 1X2×3×4 variations. And so on as far as we please.

PROB

PROBLEM IL

Any Number of different things being given; to find how many Changes can be made out of them, by taking a Given Number of Quantities at a Time.

RULE.*

TAKE a series of numbers, beginning at the number of things given, and decreasing by 1 to the number of quantities to be taken at a time, and the product of all the terms will be the answer required.

EXAMPLES.

1. How many changes may be rung with 3 bells out of 8 ?

8

7

56

6

336 the answer.

Or, 8X7X6(=3 terms) =336 the Answer.

2. How many words can be made with 5 letters of the alphabet, supposing 24 letters in all, and that a number of consonants alone will make a word. Ans. 5100480. 3. How many words can be made with 5 letters of the alphabet in each word, there being 26 letters in all, and 6 vowels, admitting that a number of consonants alone will not make a word? Ans. 137858400.

PROB.

* This Rule, expressed in algebraic terms, is as follows;

mxm-1×m-2×m-3 &c. to n terms: where m the number of things given, and n the quantities to be taken at a time.

In order to demonstrate the Rule, it will be proper to premise the following Lemma;

LEMMA. The number of changes of m things, taken n at a time, is equal to m changes of m-1 things, taken n-1 at a time.

Demonstr. Let any five quantities a b c d e be given.

First, leave out the a, and let v = the number of all the variations of every two,

bc, bd, &c. that can be taken out of the four remaining quantities b cde.

Now, let a be put in the first place of each of them, a, b, c, a, b, d, &c. and the number of changes which still remain the same; that is, the number of variations of every 3 out of the 5, a, b, c, d, e, when a is first.