be put third. This makes five times four times three ways of disposing of the first three places. For each of these there are two ways of disposing of the last two places; in all, five times four times three times two, or 120 ways of arranging the five letters. Now this method of counting the arrangements will clearly do for any number whatever of letters; so that our rule must be right for all numbers. ... We may state it in shorthand thus: the number of arrangements of n letters is 1 × 2 × 3 × ... × n; or putting dots instead of the sign of multiplication, it is 1.2.3 n. The 1 which begins is of course not wanted for the multiplication, but it is put in to include the extreme case of there being only one letter, in which case, of course, there is only one arrangement. The product 1.2.3... n, or, as we may say, the product of the first n natural numbers, occurs very often in the exact sciences. It has therefore been found convenient to have a special short sign for it, just as a parliamentary reporter has a special sign for 'the remarks which the Honourable Member has thought fit to make.' Different mathematicians, however, have used different symbols for it. The symbol n is very much used in England, but it is difficult to print. Some continental writers have used a note of admiration, thus, n! Of this it has been truly remarked that it has the air of pretending that you never saw it before. I myself prefer a symbol which has the weighty authority of Gauss, namely a Greek II (Pi), which may be taken as short for product if we like, thus, ПIn. We may now state that П11=1, П12=2, 113=6, 114=24, II5=120, II6=720, and generally that II (n + 1) = (n + 1) IIn, for the product of the first n + 1 numbers is equal to the product of the first n numbers multiplied by n+1. § 9. On a Theorem concerning any Power of a+b. We will now apply this rule to the problem of counting the terms in (a+b)"; and for clearness' sake, as usual, we will begin with a particular case, namely the case in which n=5. We know that here there is one term whose factors are all a's, and one whose factors are all b's; five terms which are the product of four a's by one b, and five which are the product of one a and four b's. It remains only to count the number of terms made by multiplying three a's by two b's, which is naturally equal to the number made by multiplying two a's by three b's. The question is, therefore, how many different arrangements can be made with three a's and two b's? Here the three a's are all alike, and the two b's are alike. To solve the problem we shall have to think of them as different; let us therefore replace them for the present by capital letters and small ones. How many different arrangements can be made with three capital letters ABC and two small ones de? In this question the capital letters are to be considered as equivalent to each other, and the small letters as equivalent to each other; so that the arrangement ABC de counts for the same arrangement as CA Bed. Every arrangement of capitals and smalls is one of a group of 6 × 2=12 equivalent arrangements; for the 3 capitals may be arranged among one another in П13,=6 ways, and the 2 smalls may be arranged in II2, 2 ways. Now it is clear that by = taking all the arrangements in respect of capital and small letters, and then permuting the capitals among themselves and the small letters among themselves, we shall get the whole number of arrangements of the five letters A B C de; namely II5 or 120. But since each arrangement in respect of capitals and smalls is here repeated twelve times, and since 12 goes into 120 ten times exactly, it appears that the number we require is ten. Or the number of arrangements of three a's and two b's is II5 divided by II3 and ПI2. The arrangements are in fact bbaaa, babaa, baaba, baaab abbaa, ababa, abaab aabba, aabab The first line has ab at the beginning, and there are four positions for the second b; the next line has a bin the second place, and there are three new positions for the other b, and so on. We might of course have arrived at the number of arrangements in this particular case by the far simpler process of direct counting, which we have used as a verification; but the advantage of our longer process is that it will give us a general formula applicable to all cases whatever. Let us stop to put on record the result just obtained; viz. we have found that (a + b) 5 = a5 + 5a1b + 10a3b2 + 10a2b3 + 5ab1 + b5. Observe that 1+5+10+10+5+1=32, that is, we have accounted for the whole of the 32 terms which would be in the last line of the tree appropriate to this case. We may now go on to the solution of our general problem. Suppose that p is the number of a's and q is the number of b's which are multiplied together in a certain term; we want to find the number of possible arrangements of these p a's and q b's. Let us replace them for the moment by p capital letters and q small ones, making p + q letters altogether. Then any arrangement of these in respect of capital letters and small ones is one of a group of equivalent arrangements got by permuting the capitals among themselves and the small letters among themselves. Now by permuting the capital letters we can make Пp arrangements, and by permuting the small letters IIq arrangements. Hence every arrangement in respect of capitals and smalls is one of a group of Пp × ПIq equivalent arrangements. Now the whole number of arrangements of the p+q letters is II (p+q); and, as we have seen, every arrangement in respect of capitals and smalls is here repeated Hp x ПIq times. Consequently the number we are in search of is got by dividing II (p+q) by Пp x IIq. This is written in the form of a fraction, thus : although it is not a fraction, for the denominator always divides the numerator exactly. In fact, it would be absurd to talk about half a quarter of a way of arranging letters. We have arrived then at this result, that the number of ways of arranging p a's and q b's is This is also (otherwise expressed) the number of ways of dividing p+q places into p of one sort and q of another; or again, it is the number of ways of selecting p things out of p+q things. Applying this now to the expression of (a+b)", we find that each of our other terms is of the form where p+q=n; and that we shall get them all by giving to q successively the values 1, 2, 3, &c., and to p the values got by subtracting these from n. For example, we shall find that The calculation of the numbers may be considerably shortened. Thus we have to divide 1.2.3.4.5.6 by 1.2.3.4; the result is of course 5.6. This has to be further divided by 2, so that we finally get 5.3 or 15. Similarly, to calculate II6 we have only to divide 4.5.6 by 1.2.3 or 6, and we get simply 4.5 or 20. To write down our expression for (a+b)" we require another piece of shorthand. We have seen that it consists of a number of terms which are all of the form but which differ from one another in having q different pairs of numbers whose sum is n. for p and Now just |