This formula comprehends all the particular cases, that can occur in any question. To find, for example, the number of arrangements, that can be formed out of m letters taken two and two, or two at a time, we make n=2, which gives we have then 2—1=1; P=m; for P will in this case be equal to the number of letters taken one at a time; there results then from this 1), m (m-2+1) or m (m for the number of arrangements taken two and two. Again, taking we find for the number of arrangements, which m letters admit of, taken three and three. Making we obtain m (m — 1) (m — 3 + 1) = m (m − 1) (m — 2.) P=m (m-1) (m-2) and n=4, m (m-1) (m2) (m — 3) for the number of arrangements, taken four and four. We may thus determine the number of arrangements, which may be formed from any number whatever of letters.* * In these arrangements it is supposed by the nature of the inquiry, that there are no repetitions of the same letter; but the theory of permutations and combinations, which is the foundation of the doctrine of chances, embraces questions in which they occur. The effect may be seen in the example we have selected, by observing, that we may write indifferently each of the 9 letters a, b, c, d, e, f, g, h, i, after the product of 6 letters abc def. Designating therefore the number of arrangements, taken 6 at a time, by P, we shall have P x 9 for the number of arrangements, taken 7 at a time. For the same reason, if P denote the number of arrangements of m letters, taken n-1 at a time, that of their arrangements, when taken n at a time, will be P m. This being admitted, as the number of arrangements of m letters, taken one at a time, is evidently m, the number of arrangements, when taken 2 and 2, will be m × m, or m2, when taken 3 and 3, the number will be m × m × m, or m3 ; and lastly, m2 will express the number of arrangements, when they are taken n and n. 139. Passing now from the number of arrangements of n letters, to that of their different products, we must find the number of arrangements, which the same product admits of. In order to this, it may be observed, that if in any of these arrangements, we put one of the letters in the first place, we may form of all the others as many permutations, as the product of n-1 letters admits of Let us take, for example, the product'a b c d e f g, composed of seven letters; we may, by putting a in the first place, write this product in as many ways, as there are arrangements in the product of six letters b cdefg; but each letter of the proposed product may be placed first. Designating then the number of arrangements, of which a product of six letters is susceptible by Q, we shall have Q7 for that of the arrangements of a product of seven letters. It follows from this, that if Q designate the number of arrangements, which may be formed from a product of n-1 letters, Qn will express the number of arrangements of a product of n letters. Any particular case is readily reduced to this formula; for making n=2, and observing, that when there is only one letter, Q1, we have 1 x 2 = 2 for the number of arrangements of a product of two letters. Again, taking Q=1x 2 and n = 3, we have 1 × 2 × 3 = 6 for the number of arrangements of a product of three letters; further, making Q=1x 2 x 3, and n = 4, there result 1 × 2 × 3 × 4, or 24 possible arrangements in a product of four letters, and so on(c). 140. What we have now said being well understood, it will be perceived, that by dividing the whole number of arrangements obtained from m letters, taken n at a time, by the number of arrangements of which the same product is susceptible, we have for a quotient the number of the different products, which are formed by taking in all possible ways n factors among these m letters. This number will therefore be expressed by P(m- ;* which being considered in connexion with * It may be observed, that if we make successively = 3, the formula n = 2, n = becomes n = 4, &c. m (m—1) m (m — 1) (m − 2) m (m − 1) (m — 2) (m3), &c. 1. 2 1.2.3 1. .2.3.4 what was laid down in art. 157, will give P(m—n+1) an xm-n for the term containing an in the development of (x+a)". It is evident, that the term which precedes this, will be express P ed by an-1xm-n+1; for in going back towards the first term, ૧ the exponent of x is increased by unity, and that of a diminished by unity; moreover P and Q are the quantities, which belong to the number n 1. =M, the two successive terms indicated P 141. If we make above, become Man-1 xm-n1 and M (m · n + 1) an xm-n n These results show how each term, in the development of (x+a)TM, is formed from the preceding. Setting out from the first term, which is xm, we arrive at the second, by making n=1; we have M1, unity for its coefficient; the result then is m since am has only 1 x m 1 а xm-1, or a xm-1. In order to pass to the third term, we make M= m a3 xm-3, and so on; whence we have the for which may be converted into this rule. To pass from one term to the following, we multiply the numerical coefficient by the exponent of x in the first, divide by the number, which marks the place of this term, increase by unity the exponent of a, and diminish by unity the exponent of x. Although we cannot determine the number of terms of this. formula without assigning a particular value to m; yet, if we numbers, which express respectively, how many combinations may be made of any number m of things, taken two and two, three and three, four and four, &c. observe the dependence of the terms upon each other, we can have no doubt respecting the law of their formation, to whatever extent the series may be carried. It will be seen that m (m — 1) (m − 2) . . . . (m — · n + 1) an xm-n 1 2 expresses the term, which has n terms before it. This last formula is called the general term of the series m (m—1) because if we make successively 1.2 a2 xm-2+ &c. n= : 1, n = 2, n=3, &c. it gives all the terms of this series. 142. Now, if (x +α) be developed, according to the rule given in the preceding article; the first term being x5 or a°x (37), 5ax*, Here the process terminates, because in passing to the following term it would be necesssary to multiply by the exponent of x in the sixth, which is zero. This may be shown by the formula; for the seventh term, having for a numerical coefficient, contains the factor m 5, which becomes 5 5 :0; and this same factor entering into cach of the subsequent terms, reduces it to nothing. Uniting the terms obtained above, we have (x+a)3 = x5 +5αx + 10a2x2 +10a3x2 + 5a*x+a*. 143. Any power whatever of any binomial may be developed by the formula given in art. 141. If it were required for example to form the sixth power of 2 x3-5 a3, we have only to substitute in the formula the powers of 2 x3 and for those of x and a; since, if we make we have -5 5 a3 respectively (2 x3 — 5 a3) = (x2 + a') ® = 4 5 +15 a' x'2 +6 a'3 x' + a' (141), and it is only necessary to substitute for x' and a' the quantities, which these letters designate. We have then or 6 (2 x3) + 6 (-5 a3) (2 x3)* +15 (— 5 a3)2 (2 x3)* 3 +20 (— 5 a3) 3 (2 x 3) 3 + 15 (— 5 a3)* (2 x 3)2 The terms produced by this development are alternately positive and negative; and it is manifest, that they will always be so, when the second term of the proposed binomial has the sign 144. The formula given in art. 141, may be so expressed as to facilitate the application of it in cases analogous to the preceding the several steps are, to form the series of numbers, |