5. Insert five harmonic means between 1 and 2-1. 11 9. The sum of three terms of a harmonical series is 12' and the first term is find the series, and continue it both ways. [10. The arithmetical mean between two numbers exceeds the geometrical by 13, and the geometrical exceeds the harmonical by 12. What are the numbers? 11. There are four numbers a, b, c, d, the first three in arithmetical, the last three in harmonical progression; shew that abc: d. [ 12. If x is the harmonic mean between m and n, shew that 13. The sum of three terms of a harmonic series is 11, and the sum of their squares is 49; find the numbers. 14. If x, y, z be the pth, qth and 7th terms of a H.P. shew that (r−q) yz + (p − r) xz+ (q − p) xy=0. 15. If the H.M. between each pair of the numbers a, b, c be in A.P., then a2, b2, c2 will be in H. P. and if the H. M. be in H. P., a, b, c will be in H.P. 16. Shew that c+2a c+26 = + 4, >7, or < 10, according as c is the A., G. or H. mean between a and b. n Things n[n. placed after each one of the n things in[w XXXIII. PERMUTATIONS. 402. THE different arrangements in respect of order of succession which can be made of a given number of things are called PERMUTATIONS. Thus if from a box of letters I select two, P and Q, I can make two permutations of them, placing P first on the left and then on the right of Q, thus: P, Q and Q, P. If I now take three letters, P, Q and R, I can make six permutations of them, thus: P, Q, R; P, R, Q, two in which P stands first. 403. In the Examples just given all the things in each case are taken together; but we may be required to find how many permutations can be made out of a number of things when a certain number only of them are taken at a time. Thus the permutations that can be formed out of the letters 404. To find the number of permutations of n different things taken r at a time. Let a, b, c, d. stand for n different things. ... First to find the number of permutations of the n things taken two at a time. ... of If a be placed before each of the other things b, c, d which the number is n-1, we shall have n-1 permutations in which a stands first, thus ab, ac, ad,. If b be placed before each of the other things a, c, d ... we shall have n-1 permutations in which b stands first, thus: : Similarly there will be n- 1 permutations in which c stands first and so of the rest. In this way we get every possible permutation of the n things taken two at a time. Hence there will be n. (n-1) permutations of n things taken two at a time. Next to find the number of permutations of the n things taken three at a time. Leaving a out, we can form (n−1). (n−2) permutations of the remaining (n−1) things taken two at a time, and if we place a before each of these permutations we shall have (n − 1). (n − 2) permutations of the n things taken three at a time in which a stands first. Similarly there will be (n-1). (n-2) permutations n things taken three at a time in which b stands first: and so for the rest. Hence the whole number of permutations of the n things taken three at a time will be n. (n − 1). (n − 2), the factors of the formula decreasing each by 1, and the figure in the last factor being 1 less than the number taken at a time. We now assume that the formula holds good for the number of permutations of n things taken r- -1 at a time, and we shall proceed to shew that it will hold good for the number of permutations of n things taken r at a time. The number of permutations of the n things taken r— - 1 at a time will be that is n. (n-1). (n-2) .....[n—{(r− 1)−1}], n.(n-1). (n-2) ...... (n−r+2). ...... Leaving a out we can form (n − 1). (n − 2) .............. (n−1−r÷2) permutations of the (n-1) remaining things taken r−1 at a time. Putting a before each of these, we shall have (n-1). (n-2)...... (n−r+1) permutations of the n things taken r at a time in which a stands first. So again we shall have (n−1). (n−2).............. (n−r+1) permutations of the n things taken r at a time in which b stands first; and so on. Hence the whole number of permutations of the n things taken r at a time will be n. (n-1). (n-2) ...... (n − r + 1). If then the formula holds good when then things are taken r-1 at a time it will hold good when they are taken at a time. But we have shewn it to hold when they are taken 3 at a time; hence it will hold when they are taken 4 at a time, and so on: therefore it is true for all integral values of r*. 405. If the n things be taken all together, r=n, and the formula gives as the number of permutations that can be formed of n different things taken all together. For brevity the formula n.(n-1). (n-2)...... 1, which is the same as 1.2.3 ...... n, is written n. This symbol is called factorial n. r is put for 1.2.3...... r ; Similarly 406. To find the number of permutations of n things taken all together when certain of the things are alike. Let the n things be represented by the letters a, b, c, d.............. and suppose that a recurs p times, * Another proof of this Theorem may be seen in Art. 480. S. A. 19 Let P represent the whole number of permutations. Then if all the p letters a were changed into p other letters, different from each other and from all the rest of the n letters, the places of these p letters in any one permutation could now be interchanged, each interchange giving rise to a new permutation, and thus from each single permutation we could form 1.2...... p permutations in all, and the whole number of permutations would be (1. 2...p) P, that is \p. P. Similarly if in addition the q letters b were changed into q letters different from each other and from all the rest of the n letters, the whole number of permutations would be 1. p. P; and if the r letters c were also similarly changed, the whole number of permutations would be r. q. p. P; and so on, if more were alike. But when the p, q and r, &c. letters have thus been changed we shall have n letters all different, and the number of permutations that can be formed of them is [n (Art. 405). Hence P.P..... = n; 1. How many permutations can be formed out of 12 things taken 2 at a time? 2. How many permutations can be formed out of 16 things taken 3 at a time? 3. How many permutations can be formed out of 20 things taken 4 at a time? 4. How many changes can be rung with 5 bells out of 8? 5. How many permutations can be made of the letters in the word Examination taken all together? 6. In how many ways can 8 men be placed side by side? |