Hence the roots are at -3, between 0 and 1, and between 2 and 3. Locate, graphically, the real roots of: 3. X3 10.x 3 4. X3 26x + 5 5. X3 15x 4 6. X3 + 2x2 + 9 0. 0. 0. 0. 7. X3 3x + 2 0. 8. x3 + 9x2 + 16x + 4 = 0. 9. x3 2x2 23x + 60 = 0. 10. x3 + 5.02 + 7x + 2 = 0. REVIEW EXERCISE XXVI Form the equation whose roots are: 1. 3 + V6 and 3 -Vā. va + 1 Vū - 1 2. and 2 X = Without solving, discuss the roots of the equation: 11. 3x2 + 17x 5 - 0. 5 + = 0. 8 1. 20. x2 – 3.0 + 5 = 0. 21. Determine the values of k that the equation (k + 2)x? – 10kx + 25 = 0 may have equal roots. 22. Form the quadratic equation, one of whose roots is 3+ V2, and the sum of whose roots is 6. 23. For what values of m are the roots of 2mx2 + 7mx x2 + 5x – 5m equal? 24. Form the quadratic equation in which one root is 3 + V2 and the product of the roots is 7. 25. Locate the roots of the equation, 23 – 3r2 – 9x + 1 : 0. 26. Form an equation whose roots shall be the cubes of the roots of the equation 2x(x – a) = a?. 27. Under what conditions will the equation x2 + p.x = 9 have roots that are reciprocals of each other? 28. Locate the roots of x3 + 4x2 – 7 = 0. CHAPTER XXVII PERMUTATIONS AND COMBINATIONS 187. All the different orders in which it is possible to arrange a given number of things, by taking either some or all at a time, are called the permutations of the things. Thus, the permutations of the figures 2 and 3, are 23, 32; the permutations of three figures, 2, 3, 4, taking two at a time, are 23, 24, 32, 42, 34, 43. All the different selections that it is possible to make from a given number of things, taking either some or all at a time, without regard to the order in which they are arranged, are called the combinations of the things. Thus, the combinations of three letters a, b, c, taken two at a time are ab, ac, bc, where the order of the letters is disregarded. Each of these combinations will furnish two permutations, as, ab, ba; ac, ca; bc, cb. When all the things are taken, there can be only one combination, as abc of the three letters a, b, c. Notation.—The symbol for the number of permutations of n things taking r at a time, s P", of n different things taking n at a time, is Pn, of 6 different things taking 3 at a time, is P. Sometimes, instead of Pn, npr or P nor is used. The symbol for the number of comb nations of n different things, taking r at a time, is Ch; of n different things, taking n at a time, is Cn. These are also sometimes written nC, or Cnor and nCnor Cnon. 188. To find the number of permutations of n different things taken r at a time. The permutations of a, b, c, taken two at a time, are ab ba ca evidently formed by writing after each of the three letters each of the other letters. Therefore the total number of permutations is 3 X 2. The number of permutations of n letters, taken 2 at a time, is found in the same way, by writing after each of the in letters, each of the remaining n - 1 letters. Therefore the total number of permutations of n letters, taken 2 at a time, is n (n − 1). If the n letters are taken 3 at a time, the permutations may be obtained by adding each of the remaining n - 2 letters to each permutation of the letters taken 2 at a time, or the total number of permutations of n letters, taken 3 at a time, is n(n − 1)(n − 2). Hence, The number of permutations of n different things, taken r at a time, is equal to the continued product of he natural numbers from n down to n - (r – 1) inclusive, the nu:nber of factors being equal to r. :.Pn = n(n − 1)(n - 2).....(n - 1 + 1) (1) To get the factors from n down to 1, we must supply the factors (n r) (n 1)(n 2).....2.1. If the right member of equation (1) is multiplied and divided by this product, which is n -r (factorial n-r means the factors from 1 to n r), we obtain, r r Equation (1) is usually more convenient in the solution of numerical problems. When all the n things are taken at a time, r = n, and equation (1) becomes, (3). Pn п EXERCISE 138 1. In how may 5 persons sit on a bench?. Substitute in the formula, P5 = ! 5 5 X 4 X 3 X 2 X 1 = 120 2. Three persons enter a public telephone station in which there are 6 empty booths. In how many ways may they select booths ? This is evidently a problem to find the number of permutations that may be made of 6 things taken 3 at a time. Hence P36 = 6 X 5 X 4 = 120. 3. How many numbers of 4 figures each can be formed with the figures 1, 2, 3, 4? 4. How many permutations may be made of the letters in the word grapes? 5. Four persons board a boat on which there are 10 vacant seats. In how many ways may they choose seats? 6. In how many different ways may 10 books be arranged on a shelf? 7. If 8 dogs compete at a dog show, in how many ways may the blue and red ribbons be awarded ? 189. To find the number of combinations of n different things taken r at a time. If from the number of permutations of n things taken r at a time, we subtract the number of permutations made from combinations by changing the order of arrangement, we should have left the actual number of combinations of the n things, taken r at a time. Thus, since two letters, as a and b, have two permutations, ab, ba, but only one combination, the number of combinations of n things, taken 2 at a time, will be only half the number of permutations of n things, taken 2 at a time. Similarly, since three letters, taken three at a time, have 3 X 2 permutations, but only 1 combination, the number of combinations of n things, taken 3 at a time, is found by dividing the number of permutations of n things, taken 3 at a time, by 3 X 2. In general then, since the number of permutations of n things taken n at a time is n, but the number of combinations only 1, The number of combinations of n different things, taken |