in clearness from its brevity. Secondly, that it is only shorthand for something which is just straightforward common sense and nothing else. We may always depend upon it that algebra, which cannot be translated into good English and sound common sense, is bad algebra. But now let us put this process into a graphical shape which will enable us to extend it. We start with two numbers, a and b, and we are to multiply each of them by a and also by b, and to add all the results. Let us put in each case the result of multiplying by a to the left, and the result of multiplying by b to the right, under the number multiplied. The process is then shown in the figure. If we now want to multiply this by a+b again, so as to make (a+b)3, we must multiply each part of the lower line by a, and also by b, and add all the results, thus: : Here we have eight terms in the result. The first and last are a3 and b3 respectively. Of the remaining six, three are baa, aba, aab, containing two a's and one b, and therefore each equal to a2b; and three are bba, bab, abb, containing one a and two b's, and therefore each equal to ab2. Thus we have :— (a+b) 3 = a3 + 3a2b + 3ab2 + b3. For example, 113 1331. Here a = 10, b = 1, and (10+1)3=103+3 × 102+3 × 10+1, for it is clear that any power of 1 is 1. We shall carry this process one step further, before making remarks which will enable us to dispense with it. In this case there are sixteen terms, the first and last being a1 and b4 respectively. Of the rest, some have three a's and one b, some two a's and two b's, and some one a and three b's. There are four of the first kind, since the b may come first, second, third, or fourth; so also there are four of the third kind, for the a occurs in each of the same four places; the remaining six are of the second kind. Thus we find that, (a+b)1=a1+4a3b+6a2b2+4ab3+ba. We might go on with this process as long as we liked, and we should get continually larger and larger trees. But it is easy to see that the process of classifying and counting the terms in the last line would become very troublesome. Let us then try to save that trouble by making some remarks upon the process. If we go down the tree last figured, from a to abaa, we shall find that the term abaa is built up from right to left as we descend. The a that we begin with is the last letter of abaa; then in descending we move to the right, and put another a before it; then we move to the left and put b before that; lastly we move to the right and put in the From this there are two conclusions to be first a. drawn. First, the terms at the end are all different; for any divergence in the path by which we descend the tree makes a difference in some letter of the result. 6 Secondly, every possible arrangement of four letters which are either a's or b's is produced. For if any such arrangement be written down, say abab, we have only to read it backwards, making a mean turn to the left' and bturn to the right,' and it will indicate the path by which we must descend the tree to find that arrangement at the end. We may put these two remarks into one by saying that every such possible arrangement is produced once and once only. Now the problem before us was to count the number of terms which have a certain number of b's in them. By the remark just made we have shown that this is the same thing as to count the number of possible arrangements having that number of b's. Consider for example the terms containing one b. When there are three letters to each term, the number of possible arrangements is 3, for the b may be first, second, or third, baa, aba, aab. So when there are four letters the number is 4, for the b may be first, second, third, or fourth; baaa, abaa, aaba, aaab. And generally it is clear that whatever be the number of letters in each term, that is also the number of places in which the b can stand. Or, to state the same thing in shorthand, if n be the number of letters, there are n terms containing one b. And then, of course, there are n terms containing one a and all the rest b's. And these are the terms which come at the beginning and end of the nth power of a+b; viz. we must have (a + b)n = a" + na"-1b + other terms + nab”-ı + br. The meaning of this shorthand is that we have n (a+b)'s multiplied together, and that the result of that multiplying is the sum of several numbers, four of which we have written down. The first is the product of n a's multiplied together, or a"; the next is n times the product of b by (n-1) a's, namely, na"-1b. The last but one is n times the product of a by (n−1) b's, namely, nab"-1; and the last is the product of n b's multiplied together, which is written b". 6 The problem that remains is to fill up this statement by finding what the other terms' are, containing each more than one a and more than one b. § 8. On the Number of Arrangements of a Group of Letters. This problem belongs to a very useful branch of applied arithmetic called the theory of permutations and combinations,' or of arrangement and selection. The theory tells us how many arrangements may be made with a given set of things, and how many selections can be made from them. One of these questions is made to depend on the other, so that there is an advantage in counting the number of arrangements first. With two letters there are clearly two arrangements, ab and ba. abc, acb, bea, bac, cab, cba, namely, two with a at the beginning, two with b at the beginning, and two with c at the beginning; three times two. It would not be much trouble to write down all the arrangements that can be made with four letters abcd. But we may count the number of them without taking that trouble; for if we write d before each of the six arrangements of abc, we shall have six arrangements of the four letters beginning with d, and these are clearly all the arrangements which can begin with d. Similarly, there must be six beginning with a, six beginning with b, and six beginning with c; in all, four times six, or twenty-four. Let us put these results together: With two letters, number of arrangements is two = 2 three times two = 6 four times three times two = 24 دو three Here we have at once a rule suggested. To find the number of arrangements which can be made with a given group of letters, multiply together the numbers two, three, four, &c., up to the number of letters in the group. We have found this rule to be right for two, three, and four letters; is it right for any number whatever of letters? We will consider the next case of five letters, and deal with it by a method which is applicable to all cases. Any one of the five letters may be placed first; there are then five ways of disposing of the first place. For each of these ways there are four ways of disposing of the second place; namely, any one of the remaining four letters may be put second. This makes five times four ways of disposing of the first two places. For each of these there are three ways of disposing of the third place, for any one of the remaining three letters may |