b=a+m' c = a + 2m' +m" d=a+3m' + 3m" +m" Here we see that the coefficients of the letters m', m", &c., in the value of any term, are the same as the coefficients of the terms in the corresponding power less one of the binomial a+x; hence the value of any term, as the nth, is 1. Find the eighth term of the series 1, 8, 27, 64, &c. We may find the letters m', m", m"", &c., by the rule given in the last section; or we may find them by subtraction: thus, 1+7×7+7X0X12 + 7X0X5X0 + IXOXEX+XO = 512 2. Find the twelfth term of the above series. 3. Find the twenty-sixth term of the series 1, 2, 3, 4, 5, 6, 7, &c. 4. Find the twelfth term of the series 1, 3, 6, 10, 15, 21, &c. SECTION CV. The Method of Differences. 1. It is required to find the sum of any number of terms of the series a, b, c, d, e, f, g, &c. If we put a = 0, the (n+1th) term of this series, will be, according to the last section, Let b=p, c=p+q, d=p+q+r, &c. Now it is evident, that the third term of the series b, c, d, &c., is the sum of three terms of the series p, q, r; since d=p+q+r; and the nth term, of the series b, c, d, &c., is the same as n terms of the series p, q, r, &c. Hence n terms of the series p, q, r, &c., is Here b, m" and m", &c., belong to the series a, b, c &c.; but b=p; and it will be seen, by taking the differences of the two series a, b, c, &c., and p, q, r, &c., that m" in the first, is just equal to m' in the second; and m" in the first, just equal to m" in the second, &c. Hence the sum of n terms of the series p, q, r, &c., is when p, q, r, &c., may represent any series, and m', m", m"", &c., the first terms of its successive orders of differences, and the above formula will represent the sum of n terms of any serios. 2. Find the sum of 10 terms of the series 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, &c. 1, 1, 1, 1, 0, 0, 0, p = 1, m' = 1, m" = 0, n= 10, and the formula becomes 10+45= = 55 3. Find the sum of 20 terms of the series 12, 22, 32, 42, 52, 62, 72, &c. 4. Find the sum of 12 terms of the series 1, 2, 3, 4, 5, 63, &c. In arsenals, it is customary to pile up balls in the form of pyramids. A square of balls is placed upon the ground, upon this another square of balls is placed, the side of the second square containing one ball less than that of the first; then a third square is placed on the second, &c.; each succeeding square of balls having its side one less than the preceding. The last square contains four balls; and upon this square one ball is placed, which completes the pyramid. 5. How many balls are there in a pyramid, having 20 balls in one side of its base? 6. How many balls are there in such a pile, supposing it only completed to of its height? 7. Supposing the base of the pyramid to be a rectangle; one side of which is 12 balls, and the other 10; how many balls are there in the whole pyramid? 8. How many balls would there be, if the base were a right parallelogram, one side of which contained 20 balls, and the adjacent side 36? SECTION CVI. Permutation. We can write the two letters, a and b, in two different positions with respect to each other: thus, we can write them 2 2 ab, or ba If we have another letter c, to write along with a and b, we can place it on the left of ab, between a and b, or on the right, and the three letters would have three different positions with respect to each other: viz. 2 cab, acb, abc We can also place c on the left of, ba, between b and a, or on the right, and we have 5 cba, bca, bac If, then, we have three letters, we can arrange them six different ways. If we have another letter d, it is evident we can place it in four different places, in each one of the six arrangements of the three letters a, b and c. Hence, we would have four times six different arrangements of the four letters a, b, c, d. If we had another letter e, we might place it in 5 different positions in each one of the 24 arrangements of the 4 letters a, b, c, d, &c. If, then, it be required to find how many different ways we can arrange n letters, or things, we may find it by multiplying the numbers 1, 2, 3 together, until we arrive at the number n, inclusive; the last product is the number required. 1 X 2 X3 X 4... X n = number required. 1. How many different ways may 10 different letters of the alphabet be arranged? 2. The roll of a class contains 12 names. How many different ways may these names be arranged? 3. How many different whole numbers may be expressed by the 9 digits, the whole of them being used to express each number? SECTION CVII. Permutations. To find the number of changes which can be made of m things, taken n at a time. 1. The number of changes which can be made of 7 things, taken 1 at a time, is evidently 7. If we take each one of the remaining letters, and combine it with the one taken, we shall have as many times 7 changes, as there are letters remaining; that is, 7 × 6. These are all the changes which can be made, taking 2 at a time. After taking two at a time, we may take any one of the permutations of 2, and place along with it one of the remaining letters, and we shall have as many times 7 × 6 per nutations of 3 as there are letters remaining; and these are all the permutations of 3. We can proceed in the same way with any number at a time. Let m be the number of things, and n the number taken at a time; then we have m (m — 1) (m — 2) (m the number -3,) &c., to n factors of changes required. 2. How many arrangements of 5 letters can be made out of 12? 3. How many different words can be spelled with 20 letters three letters in a word? 4. There are 14 names in a hat; from which a person is to draw out 3 at a time, 1 by 1. How many different ways may he chance to draw them? 5. How many different numbers may be expressed with the 9 digits, all these numbers being over 10, and under 100? 6. How many that are over 100 and under 1000? SECTION CVIII. Different Combinations. 1. There are given a number of letters, a, b, c, d, &c.; and, it is required to find how many different combinations of 2, 3 or 4, &c. can be made out of a, b, c, d, &c.; so that no two combinations of letters shall contain all the same letters: thus, all such combinations of 2 out of the 3 letters a, b, c, are ab, a c and b c. Now we say, that if m be the number of letters, and n the number to be taken at a time, the whole number of such combinations will be expressed by the following formula: 2. For if there be 2 letters, a and b, then there can be only one combination of 2 letters: viz. ab. If there be another letter c, it may be combined with each of the other letters, and we shall have 1+2, combinations, of 2 out of the 3, a, b, c. If we had another letter d, we might make 3 other combinations of 2 letters by combining d with each one of the other letters a, b, c. Then the whole num |