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common logarithms of all numbers may be readily found.

139. The connexion between common and Napierian logarithms is seen from (138), where it has been shown (5) that if N be any number then

loga N log N :: log, blog, a.

Let a 10, and be; then will

logo N log, N:: log, e: log, 10 :: 1:2·3025851;

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the Napierian log N = 2·3025851

X common log N.

We shall now work one example at length as a pattern for the computation of logarithms to any base.

Ex. Find the common logarithm of 2.

By the series (12) we have, when n = 1 and a = 10,

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1. Find the common logarithms of 5, 12, 18, 22.5 and 00072, having given log10 23010300 and log10 3 = 4771213.

· Ans. ·6989700, 1.0791812, 1·2552725, 1·3521825, and 4·8573325. 2. Given logo 3 = 4771213 and logo 7: logarithms of 021 and 1470.

= 8450980, to find the Ans. 2 3222193 and 31673173. Ans. 1 0413927.

3. Calculate the common logarithm of 11.

4. Given the logarithms of 2 and 3, as in Example 1, to find the loga

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Ans. 2 log 3 4 log 2 1 7501225;

and 5 log 2

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5. Find the logarithm of 180 to the base 6, or solve the equation

6* = 180.

log 3 3=2·0280287.

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7. Find the value of x in the equation ab*= c.

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140. The permutations of any number of quantities are the different arrangements which can be made with them, when taken two together, three together, etc., or all together: thus the permutations of a, b, c, when taken three together, are a bc, acb, bac, bca, cab, cba. The combinations of any number of quantities are the different collections which can be made of any assigned number of them, without reference to the order of their arrangement: thus, abc, abd, acd, and bed are different combinations of the four letters a, b, c, d, taken three together. 141. The number of permutations of n different quantities, taken r together, is

{ n − (r− 1)}.

n (n 1) (n − 2) For if a, b, c, d, etc., be the n quantities, then, if taken two together, a may be placed before each of the (n - 1) quantities, and thus we have (n-1) permutations. If b be placed first, we shall have also (n - 1) permutations, and c placed first will give (n - 1) permutations, and so on throughout the n different quantities: hence the total number of permutations, two together, is n (n − 1).

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If the n quantities be taken three together, then, omitting any one of then quantities, as a, the number of permutations of the remaining (n - 1) quantities is (n - 1) (n − 2) by the preceding case when taken two together. Place a before each of these permutations, two together; then we shall have (n − 1) (n − 2) permutations taken three together, in which a stands first; and there must be the same number of permutations where b, c, d, etc., successively occupy the first place in each; hence the entire number of permutations of the n different quantities must be n times the number (n - 1) (n − 2), or n (n − 1) (n − 2).

--

In a similar manner, the number of permutations of n quantities taken four together is n (n − 1) (n − 2) (n − 3), and generally when taken r together, the number is

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If the quantities are taken all together, then r = n, and the number of permutations is

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3.2.1, or 1.2.3.4.

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142. The number of permutations of n quantities taken all together, of which p quantities are alike, q other quantities are alike, r other quantities are alike, and so on, is

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For if the p quantities were unlike, they would form 1.2.3. . Р permutations, instead of only one when they are alike; therefore the whole number of permutations must be diminished 1.2.3. . . Р times, if p quantities are alike, hence the number of permutations, if p quantities are alike, is

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For a similar reason, the whole number of permutations, if q other quantities are alike, must be diminished 1.2.3. expression for the number of permutations becomes

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If r other quantities are alike, the expression is

(1.2.3.

1.2.3.4.

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and so on; hence the truth of the proposition is established.

143. The number of combinations of n different quantities, taken r together, is

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For the number of permutations of n quantities, taken two together, is n (n − 1), and there are two permutations, ab, ba, corresponding to one combination, ab; hence there are twice as many permutations as n (n − 1) combinations; and therefore the number of combinations is

1.2

If the quantities are taken three together, the number of permutations is n (n - 1) (n − 2), and there are 1.2.3 permutations for each combination; therefore there are 1.2.3 times as many permutations as n (n − 1) (n − 2) combinations, and the number of combinations is

-

1.2.3

In a similar manner it may be shown that the number of combinations of n quantities, taken r together, is

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144. The number of combinations of n things, taken r together, is equal to the number of combinations taken n - r together.

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r for r in the last result, the number of combina

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but these results (a) and (B) are equal to each other, for

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(n -- 2) (n − 1) n

3.2.1

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as is evident by clearing the equation of fractions; for then both sides are identical, but written in reverse order. Thus if n 8, and r = 3, 8.7.6 4.5.6.7.8 or the number of combinations of 8 quan

then

=

1.2.3 5.4.3.2.1'

tities, taken 3 together, is equal to the number of combinations taken 5 together, as is evident by bringing these fractions to a common denominator.

These combinations of n quantities, when taken r together, and n − r together, are said to be supplementary to each other.

145. Also if C, denote the number of combinations of n quantities taken r together, then the sum of all the combinations that can be made of n quantities, taken one, two, three, ... n together is

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1. Find the number of permutations of the letters in the word change.

Ans. 720.

2. Find the number of permutations of the letters in the word appliAns. 4989600.

cation.

3. In how many ways may 7 persons seat themselves at table?

4. How many changes may be rung many with the whole peal?

Ans. 5040. with 5 bells out of 8, and how Ans. 6720 and 40320.

5. The number of permutations of n quantities three together, is to the number of permutations five together, as 1 to 42; find n.

Ans. n = 10.

6. The number of combinations of n quantities four together, is to the number two together, as 15 to 2; find n. Ans. n = 12.

7. On how many nights may a different guard be posted of 4 men out of a company of 36? and on how many of these will any particular soldier be on guard? Ans. 58905 and 6545.

8. If a company consisting of 30 men are drawn up in column, with how many different fronts can that be done when 5 men are always in front? Ans. 142506.

9. A captain, who had been successful in war, was asked what reward he expected for his meritorious services; he replied that he would be satisfied with a farthing for every different file of 6 men he could form with his company, which consisted of 100 men; what was the amount of his request? Ans. 12417211. 5s.

4

INTEREST AND ANNUITIES.

146. To find the amount of a given sum of money in any number of years, at simple interest.

Let P denote the principal or sum at interest in pounds, r the rate of interest of 17. for one year, t the number of years the principal is at interest, and A the amount of principal and interest at the end of t years; then since

r

the interest of 11. for one year, tr the interest of 11. for t years, Ptr the interest of Pl. for t years,

=

.'. A=P+Ptr = P(1 + tr)

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From this equation, any three of the quantities P, r, t, A being given, the fourth may be found.

147. To find the amount of a given sum of money in any number of years, at compound interest.

In addition to the notation in the former proposition, we shall make use of R to denote the amount of 17. for one year; then R = 1 + r. Now P in one year will amount to P+Pr = P (1 + r) = PR, and this being the principal for the second year, the corresponding amount will be PR + PR r = PR (1 + r) = P R2. In a similar manner PR is the amount in three years, and consequently in t years the amount will be

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(2),

or, log A= log P+t log R. . . (2′).

Cor. If the interest is paid half-yearly, then 2 t will be the number of payments, and the rate of interest; hence we have in this case

2

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Hence we can find the time in which any sum at compound interest will amount to twice, thrice, or m times itself.

Thus if A 2 P; then by (2) we have 2

if A or, if A

=

R', and t log R = log 2; 3P; then 3 R', and t log R = log 3; = m P; then m = R', and t log R = log m.

=

=

EXAMPLES.

1. If 5007. be allowed to accumulate at compound interest at the rate of 57. per cent. per annum, what will be the amount at the end of 21 years? Ans. 13921. 19s. 74d. 2. In what time will any sum of money double itself at the rate of 4 per cent. per annum, compound interest being allowed?

Ans. 15 7473 years.

3. If the population of a city contain one million of inhabitants, and increase at the average annual rate of 3 per cent., what will the population amount to at the end of 10 years? Ans. 1343915 inhabitants.

ANNUITIES CERTAIN.

148. An annuity is a sum of money which is payable at equal intervals of time.

When the possession of an annuity is not to be entered upon until the

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