To find the number of permutations of three things a, b, c, it must be observed that a may remain first, while bc change; thus we have a b c, a cb; likewise b and c may remain first while the others change, so we have 1×2×3=6 permutations.

In the permutations of four things, we notice likewise that every one thing may be first, while the other three change; and as three things give 1×2×3 permutations, we shall have 1×2 x3x4, or 24 permutations of four things.

Similarly with five things; every one may be first, while the four others change. Thus we obtain as the number of permutations of five things :

1×2×3×4×5 or 120.

The permutations of six things, taken all together, are:

1×2×3×4×5×6=720.

The law which is here perceived may be generalised thus:

To find the number of permutations of n things, taken all together, we have 1× 2 × 3 × 4 × ......(n—3) (n—2) (n−1)n for the required number.

Ex. 1. Required the number of different ways in which 10 persons can be placed round a table.

Ex. 2. How many changes can be rung on 12 bells? and how long would they take in ringing once over, supposing 10 changes might be rung in a minute, and the year to consist of 365 days. Ex. 3. How many different ways can the 7 notes in music be

varied?

374. Let us now determine how many permutations can be found out of a certain number of things, by taking two and two, three and three together, &c. We shall find that the premutations of six things, a, b, c, d, e, f, taken two together, are with a as first letter ab, ac, ad, ae, af,

and so on; and the whole number is 6×5=30.

When three are taken together:

ab, as first, gives

ca, cb, cd, ce, cf,

abc, abd, abe, abf,

Thus every permutation of two letters gives 4 of three letters, and as there are 6×5 changes of two letters, we have 6×5×4 or 120 permutations of three letters.

Then each permutation of three gives 3 changes, and 6× 5 × 4 ×3 or 360 expresses the number of permutations of six things, four together.

Similarly the number of permutations of five things out of six is 6 × 5 × 4×3×2 or 720.

It follows that if 12 different things be given, out of which 6 are taken, the number of changes they form is 12× 11 × 10×9 8x7, or 6652880.

Then generally the number of permutations of n things, taken Two together, are n (n—1)

Three

Fonr

66

Five

66

n (n—1) (n—2)

n (n−1) (n—2) (n—3)

n (n-1) (n-2) (n—3) (n—4)

n (n-1) (n-2) (n-3) (n—4)......(n—p+1).

The pupil will do well to express this law in correct language. Ex. 1. How many changes can be made by taking 4 digits out of the nine?

Ex. 2. How many changes can be rung on 6 bells, out of 12 ?

Ex. 3. How many permutations can be made with 4 letters out of the 26 which compose the alphabet?

375. We shall now inquire how the permutations decrease, when some of the things have the same letter. Suppose two letters are given, and both alike, then the two changes are 1 x 2 reduced to one; therefore, or 1, is equal to the expression 1x2'

when both letters are the same.

If three letters be alike, the

or 1,

six permutations are reduced to one; therefore,

1×2×3

1x2x3'

is the expression of the changes when the three letters are the

Now, to ascertain the permutations of aaabbc, we have to observe that six letters, if they are all different, would give 1 × 2 × 3 × 4×5×6 changes; but since a occurs three times, these changes must be divided by 1×2×3; also, because b

occurs twice, we must divide by 1x2. Hence the required 1x2x3x4x5x6

number of permutations is

or 60.

Let a be repeated p times, b repeated q times, c repeated r times, &c., and we shall then have the general expression : 1×2×3× .p× 1 × 2 × 3 × ...q×1×2×3×...r for the per

1 × 2 × 3 × 4×.........(n—3) (n−2) (n−1)n mutations of any number of letters, when there are p of any sort q of another, r of another, &c.

Ex. 1. How many permutations can be made of the letters in the word commemorate?

The whole number of letters is 11; o is repeated twice, m three times, and e twice. .. The permutations are:

1×2×3×4 × 5 × 6 × 7 × 8 × 9 × 10×11

1 × 2 × 1 × 2×3×1×2

-1663200.

Ex. 2. How many permutations can be made of 4 oranges, 4 apples, and 4 pears?

Ex. 3. Required the number of permutations that can be formed out of the letters of the word "Disinterestedness."

Ex. 4. How many different numbers can be made out of the following digits: 1223334444?

376. Now, if instead of finding how many changes and quantities a, b, c, d, e may undergo, we find the different groups that can be formed out of them, so that no two groups consist of the same things, we shall determine the combinations of those quantities. Thus ab, ac, ad, ae, bc, bd, be, cd, ce, de, are the combinations of five things, taken two and two

abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde, are the combinations of five things, taken three and three.

abcd, abce, abde, acde, bcde, are the combinations of five things, taken four and four.

abcde is the only combination of five things taken altogether. We have seen that the number of permutations of five things taken together is 5×4, and each combination admits of two 5 × 4 permutations; therefore,

expresses the number of com

binations of five things, taken three together.

We have already found that n(n-1) (n-2) (n-3)...(n-p+1) express the permutations of n things, taken p and p together; hence the number of combinations of n things taken, also p and p together, is:

n(n—1) (n—2) (n—3).........................(n—p+1)

Ex. 1. In how many ways can four soldiers be taken to mount guard from a company of 50 men?

The number of different guards is evidently the same as the number of combinations of 50 things, taken four at a time.

Ex. 2 A groom is ordered to bring 4 horses from a stable containing twelve horses, but having forgotten which were wanted, he brings 4 taken at random. What chance has he of being right?

Ex. 3. How many different three-coloured flags can be made with the primitive colours, viz., red, orange, yellow, green, blue, indigo, and violet?

Ex. 4. How many signals may be made by a telegraph consisting of six boards, each making one motion?

Ex. 5. There are four companies, in one of which there are 18 men, in another 20, and in each of the other 24 men. What are the combinations that can be made with 4 men, one out of each company?

In this problem let us establish the principle on which we should proceed. If there were two companies containing 18 and 20 men respectively, and every man of one company is combined with every man of the other, the number of combinations is evidently the product of the two numbers of men, or 18 × 20. Again if there be another company of 24 men introduced, each of these being combined with 18 × 20 combinations, will make 18x 20 x 24 combinations of three companies, one man being taken out of each company. The same method of reasoning will be true for any number of companies. Therefore the number of combinations required is the product of the numbers which express the number of men in each company. If there were the same number of men in each company, that number raised to the power expressed by the number of companies would be the

answer.

Hence, generally, if n represent the number of companies containing respectively, p, q, r, &c., men, and one man being taken out of each company, the number of combinations is pxqxrx &c., viz. the continued product of the numbers expressing the number of men in each company. If there be the same number of men in each company, then p" is the number of combinations. Thus in our question the answer is 18 × 20 × 24 × 24 or 207360.

Ex. 6. In a school of 106 boys, it is found that 12 of them are in room No. 1; 16 in room No. 2; 18 in room No 3; and the others are equally divided in rooms Nos. 4, 5, and 6. How many ways may 6 pupils be taken, one out of each room?

Ex. 7. Required the number of ways in which 8 men, 6 women, and 5 boys can be taken, so as always to have one out of each set?

Ex. 8. How many combinations are there in throwing 4 dice? Ex. 9. Find the number of different triangles into which a polygon of 12 sides may be divided by joining the angular points.

Ex. 10. How many different sums may be formed with a sovereign, a half-sovereign, a crown, a half-crown, a shilling, a sixpence, a fourpence, and a penny?

Ex. 11. How many words can be found, consisting of 3 consonants and a vowel, from an alphabet which consists of 21 consonants and 5 vowels?

Ex. 12. In a box of figures of animals and riders every figure is divided into four parts. Now taking four parts together, viz., any head and neck, stomach and legs, hind quarters, and rider, they form a complete figure. There are 12 sets. Find the number of combinations.

LOGARITHMS.

377. If there be written the natural series of numbers which form an arithmetical progression, the difference of which is 1; and also different geometrical progressions, the ratio being 2, 3, 4. 5, &c.......10, &c.; and placing these series under one another, as follows,