n 11.10.9.8.7.6.5.4.3.2.1 =831600. 1.2x1.2.3.4 2. How many different numbers can be made of the following digits ; 1, 2, 2, 7, 7, 7, 5, 5, 5, 5 ? Ans. 12600. 3. How many permutations may be made of the letters in the word “examination." Ans. 4989600. Theorem 3. The number of combinations that can be formed out of (n) things, taken two and two together, is i{n.(n · 1)}; (-1):(12) when taken three and three together 1.2.3 Now since each combination ab admits of two permutations, .. when the quantities are taken two at a time, there will be 1.2 times as many permutations as combinations. Now by Theorem 1, the number of permutations of (n) things, taken two at a time, =n.(n-1); hence the number of permutations = 1.2 times the number of combinations; .. the number of combinations 7.(n-1) 1.2 Again, by Theorem 1, there are n.(n-1).(n - 2) permutations in (n) things, taken three at a time ; but each combination of 3 things admits of 1.2.3 permutations, .'. n.(n-1).(n--2) = 1.2.3 times the number of combinations, i. e. the number of combiną. 7.(n-1). (m2) tions By following the same method, it ap 1.2.3 pears that the number of permutations of (n) quantities, taken (r) at a time,=n(n-1).(n-2)...(n-r+1). Also each combination of (r) things admits of 1.3.3.4. ...r permutations ; : the number of .. permutations = 1.2.3.... times the number of combinations ; hence the number of combinations in (n) quantities, taking r of them at a time, n.(n-1).(n-2)....(n-r+1). 1.2.3. Theorem 4. The total number of combinations which can be formed of (n) things taken, one at a time, two at a time, &c. = 21-1. First let the quantities be taken one at a time, then the number of combinations is ...nn. If taken two at a time, the number is -1 (n-1). (1-2) If taken 3 at a time, the number is n. 2 1.2.3 (1-1).(n-2)/(-3) If taken four at a time, the number is n. 1.2.3.4 ... the total number an+ n.(n-1) %.(n-1).(7-9) +-&c. 1.2 1.2.3 =(1+1)"—1=2" — 1. n . ...? n "3 . + On Logarithms. 123. Logarithms are the powers to which a given number must be raised, in order to become equal to other given numbers. We know that 10=10, 10=100, 103 1000, &c.; therefore, if 10 were the given number, 1, 2, 3, &c. are the powers to which 10 must be raised, in order to become equal to 10, 100, 1000, &c. Now, suppose it were required to find a number (é) such, that 10 =5; it is evident that is not so great as 1, because 10 = 10: } 13 suppose is, then 10 =100% = a number less than 5. a more than or .666666; but if x were = .69897, then 10- 2 very nearly; this number .69897 is the log. of 5. It is likewise found that 10.30 3 .. x is .301030 10.477121 10.698970 1.94139211, &c. i. e. .301030 is the log. of 2; .477121 is the log. of 3; .698970 is the log. of 5, and 1.041392 is the log. of 11, &c. Since 10 is the base of our present system of arithmetic, the same number has been chosen for the base of the logarithmic system now in use. It is evident that any other base (a) might be taken, and then it would be required to find all the values of %, which would make a : each of the different numbers whose lo-, garithms we wish to find. Let N, n be any two numbers whatever; and suppose a= N, then 2 is the log. of N. And ał = n, then x is the log. of n. Also-NXn=aixa=aits. Now (c+ą) is the log, of NXn by the definition ; Log. N+log. n=log. (NXn). Hence the log. of the product of two numbers = the sum of the logarithms of the numbers. Thus, log. 3X5=log. 3+ log. 5; Log.(9X8X7)=log.(978)+log.7=log.8+log.9+log.7. N .. Log. => by the definition of loaz garithms, =log.N-log.n. 124. Hence the logarithm of the quotient of any two numbers is equal to the difference of the logarithm of those numbers. If N be less than n, x is less than x; .. therefore the logarithms of any proper fraction is negative. Since Naa'; ::: N"=a" and the log. of N=mx=mlog. N. 1 1 Also N = a; ... Log. N" *2=log.N. 125. The properties of logarithms here given are true in every n n mx n n n system ; but we have selected the system whose base is 10, because it is the most convenient for all purposes of calculation. It is also evident, that if a series of numbers be taken in geometrical progression, their logarithms will be in arithmetic progression. Take the common scale 10-5, 106, 103, 107, 104, 1, 10', 102, 103, 104, &c. 1 1 Or ...1, 10, 100, 1000, &c. 100000' 10000" Or .00001, .0001, .001, .01, .1, 1, 10, 100, 1000, &c.; each of these series forms a geometric progression, and the logarithms of the terms of the series are —5, 4, -3, -2, -1, 0, 1, 2, 3, &c. which is an arithmetic progression. And in general, aʻ, att, at+4, az+3, &c. form a geometric series of numbers whose logarithms are x, x+r, x+2r+&c. in arithmetic progression. N 101 N 105 -10%-2, 102 103 &c. hence, by dividing any number by 10, its logarithm is diminished by 1; dividing by 100, diminishes the logarithms by 2, &c. therefore taking any number, 3854, whose logarithm is 3.5859117, The log of 385.4 =2.5859117 38.54 =1.5859117 .003854—3.5859117 &c. Let a=N; then if x be found in terms of N and a, it will be the logarithm of N to the base a. Suppose N=1+n a=1+6 Then (1+6)=l+n, and (1+6)=(1+n)' 6=1== +&c. 72 13 Now (1+6)=173 (+ -2 1.2.3 9 kg 1 1,1 n? 1,1 1 no Also (1+n)'=l+n+:{ -1}; til -2}. 1.2 1.2.3 1 Now suppose y to become very great; then and become y y very small when compared with -1, -2, &c. If y be increased without limit we obtain 1 {6–38+388–46*+&c.}={n—no+fn —4n* &c.} }; -1n y y 14n+ine_4n+&c. =M:{n-fn?+$no-m (aman 1)-(2-1)+ (0-1)&c. 1). 1.2 + -1}+ Y 2 -n &c.} &c.} 1-n ; then +5(N=134&c.} +&c.}. If M= the modu(a-1)-1(2-1)+ (0-1) - &c. lus of the system. Hence log. N ; or log. (1+n) M{n-fn-In-4n+ &c}. If n be any whole number, the terms of this series inerease, and .:the series itself does not converge. To obtain a converging series we have by a similar method, log. (1-n)=M-nen-in-n_ Subtract this series from the other, and 1+n &c. itu N 2 1 1 .. Log. =2M + &. N_2 3.(N-1)3 1 (N—2); an expression which converges with sufficient rapidity. . 1 N-) 2N_1 (2N – 1187&c.}; two terms of į which series will give the logarithms to 14 places of decimals, whenever N is = 100 or upwards. 1 This series is found by supposing na 2N 1 Prob. 2. To construct a Table of Logarithms. Since a may be arbitrarily assumed, let it be so taken that (a--1)-1(a-1)+F(a–1&c=1 Then the first number whose logarithm we must find is 2. Now log. 4= log.2=2 log.2. 1 1 .. Log. 4 or 2 log. 2-2 21十年十。 1 1 1 .: Log. 2 24+ = .693147 3 (1 1 1 Log. 3 2 + =1.098612 2 3,23 Log. 4 = 2 log. 2... =1.386294 1 1 Log. 5 { + 4 5.45 Log. 6 log. 3 + log. 2.. =1.791759 * See Peacock's examples in the diff. and int. calculus, p. 61. ... Log.N=2M (N-1 +3-(N_]) + +5°N —13*+&c.} + log. It may be easily shewn that log; N + 1) = 2 log. N-— log . (N—1) —2M {2+ + 5.35+ &c.} +log. 2. + list 3+5.364&c.} 5.25 +&c.} 2{6+ + 2.3.4(logN)+&e.* 1 1 1 Log. 7 6.65 +&c.} +log.5=1.245892 6 3.63 Log. 8 log 4 + log 2 or = 3 log 2 ...=2.079441 Log. 9 log 4 2 lgo 3.. =2.197224 Log.10 = log 5 + log 2 log 2.302585 Log.11 = 2 = 2} 10 +3.10+5.10+&c.} + log 9=&c. 't * The logarithms here calculated are called Napier's logarithms, from the name of their noble and ingenious inventor ; but are commonly known by the name of hyperbolic logarithms, because they can be represented by the areas of the equilateral hyperbola, contained between its assymptotes. To find the value of the base (a) of this system, whose modulus equals 1. Since log N, or log (1+n)=n- n+1n—*+&c; therefore by reversion of series, 1 1 NLog N+i(logN)'+ -(logN)' + 2.3 4 1 Nl+ 1 1+ + +1.2.3.4+&c. =2.7182818 = the number whose hyperbolic log is 1, or = the base of the system. According to this system (2.7182818).693147=2 (2.71828181.98812)=3, &c. Again, since Hyp log 2.71828181 therefore Hyp log (2.7182818)=2 Hyp log (2.7182818)=3, &c. Hence in this system the numbers whose logarithms are 1, 2, 3, &c. are decimal numbers, and therefore very inconvenient for the ordinary purposes of calculation. To change hyperbolic logarithms into common logarithms, or those generally used. In this system the base (a)=10, and its log=1. Now in the system whose modulus M, Log N=M{n-1n+in-&c. and in the Naperian system, Hyp log N=n=n?+in-f-c. because M=1. Therefore N=MXHyp log N; and if N=10, we have 1 1 1-MX Hyp log 10; ... M= =:43429448 Hyp log 10 2.305585 and common log N=43429448X Hyp log N. 11 1 1 34 |