2. Take the series 1, 2, 3, 4, &c. up to the number of given things of the first sort, and the series 1, 2, 3, 4, &c. up to the number of given things of the second sort, &c.

3. Divide the product of all the terms of the first series by the joint product of all the terms of the remaining ones, and the quotient will be the answer required.

EXAMPLES.

1. How many variations may be made of the letters in the word Bacchanalia?

1x2(=number of cs)= 2 1x2×3×4(=number of as)=24

1×2×3×4x5×6×7×8×9×10×11(=number of letters in the

word)=39916800.

2x24=48)39916800(831600 the answer.

151

76 288

2. How many different numbers can be made of the foldowing figures, 1220005555?

Ans. 12600.

3. What is the variety in the succession of the following musical notes, fa, fa, fa, sol, sol, la, mi, fa?

Ans. 3360.

if two of the quantities be alike, or the 4 quantities be aabc, the number of variations will be reduced to 12, which may be ex1×2×3×4 1x2

pressed by

= 12.

And by reasoning in the same manner it will appear, that the number of changes, which can be made of the quantities, abbccc, 1x2x3x4x5X6 1X2X1X2X3

is equal to 60, which may be expressed by

60; and so of any other quantities whatever.

PROBLEM 4.

To find the changes of any given number of things, taken a given number at a time; in which there are several given things of one sort, several of another, &c.

RULE.*

1. Find all the different forms of combination of all the given things, taken as many at a time as in the question.

2. Find the number of changes in any form, and multiply it by the number of combinations in that form.

3. Do the same for every distinct form; and the sum of all the products will give the whole number of changes required.

NOTE. To find the different forms of combination pro

ceed thus:

1. Place the things so, that the greatest indices may be first, and the rest in order.

2. Begin with the first letter, and join it to the second, third, fourth, &c. to the last.

3. Then take the second letter, and join it to the third, fourth, &c. to the last; and so on through the whole, always remembering to reject such combinations as have occurred before; and this will give the combinations of all the twos.

4. Join the first letter to every one of the twos following it; and the second, third, &c. as before; and it will give the combinations of all the threes.

* The reason of this rule is plain from what has been shown before, and the nature of the problem.

5. Proceed in the same manner to get the combinations of all the fours, &c. and you will at last get all the several forms of combination, and the number in each form.

EXAMPLES.

1. How many changes may be made of every 4 letters, that can be taken out of these 6, aaabbc ?

2. How many chunges can be made of every 8 letters out of these 10, aaaabbccde? Ans. 22260.

3. How many different numbers can be made out of 1 unit, 2 twos, 3 threes, 4 fours, and 5 fives, taken 5 at a time? Ans. 2111.

PROBLEM 5.

To find the number of combinations of any given number of things, all different from one another, taken any given number at a time.

RULE.

1. Take the series 1, 2, 3, 4, &c. up to the number to be taken at a time, and find the product of all the terms.

m M

* This rule, expressed algebraically, is ×1× X

2

M-2

X

&c. to n terms; where m is the number of given quan

tities, and n those to be taken at a time.

DEMONSTRATION OF THE RULE. 1. Let the number of things to be taken at a time be 2, and the things to be combined

=m.

Now, when m, or the number of things to be combined, is only two, as a and b, it is evident, that there can be only one combination, as ab; but if m be increased by 1, or the letters to be combined be 3, as abc, then it is plain, that the number of combinations will be increased by 2, since with each of the former letters, a and b, the new letter c may be joined. It is evident therefore, that the whole number of combinations, in this case, will be truly expressed by 1+2.

Again, if m be increased by one letter more, or the whole number of letters be four, as abcd; then it will appear, that the whole number of combinations must be increased by 3, since with each of the preceding letters the new letter d may be combined. The combinations therefore, in this case, will be truly expressed by 1+2+3.

In the same manner it may be shown, that the whole number

2. Take a series of as many terms, decreasing by 1 from the given number, out of which the election is to be made, and find the product of all the terms.

of combinations of 2, in 5 things, will be 1+2+3+4; of 2 in 6 things, 1+2+3+4+5; and of 2, in 7, 1+2+3+4+5+6, &c.

Whence universally, the number of combinations of m things, taken 2 by 2, is =1+2+3+4+5+6, &c. to m―1 terms.

But the sum of this series is

m m- 1
+
which is the same
"
2

as the rule.

2. Let now the number of quantities in each combination be supposed to be three.

Then it is plain, that when m=3, or the things to be combined are abc, there can be only one combination; but if m be increased by 1, or the things to be combined be 4, as abcd, then will the number of combinations be increased by 3; since 3 is the number of combinations of 2 in all the preceding letters abc, and with each two of these the new letter d may be combined.

The number of combinations therefore, in this case, is 1+3.

Again, if m be increased by one more, or the number of letters be supposed 5; then the former number of combinations will be increased by 6; that is, by all the combinations of 2 in the 4 preceding letters, abcd; since, as before, with each two of these the new letter e may be combined.

The number of combinations therefore, in this case, is 1+3 +6.

Whence universally, the number of combinations of m things, taken 3 by 3, is 1+3+6+10, &c. to m-2 terms.