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Hence the roots are at -3, between 0 and 1, and between 2 and 3.

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Without solving, discuss the roots of the equation:

11. 3x2 + 173 – 5 = 0.
12. 10x2 – 21x – 10 = 0.
13. 25x2 = 10x – 1.
14. 3.2 – 5x + 4 = 0.

7 155
9. + 2 + x – 8.5

-= 0.
16. 4x2 – 7x = 10.
17. 2x2 – 3 + 5 = 0.
18. 2x - x = 1.
19. 7x2 + 13x = 5.

20. x2 – 3x + 5 = 0. 21. Determine the values of k that the equation (k + 2)x2 – 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 – 3x2 – 9x + 1 = 0.

26. Form an equation whose roots shall be the cubes of the roots of the equation 2x(x – a) = al.

27. Under what conditions will the equation x2 + px = q 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, be, 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.

Votation.—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 Pg. Sometimes, instead of P, P, or Px, is used.

The symbol for the number of combinations of n different things, taking r at a time, is C); of n different things, taking n at a time, is Cn.

These are also sometimes written "f, or Cruz and of, or Com

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

ah ha ca ac be cb

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 – (1 – 1) inclusive, the nu:nber of factors being equal to r.

P = n(n − 1)(n − 2).....(n ” + 1) (1) To get the factors from n down to 1, we must supply the factors (n r)(n 7 – 1)(n — 2).....2.1.

If the right member of equation (1) is multiplied and divided by this product, which is in -r (factorial n r means the factors from 1 to n r), we obtain,

'n

(2).

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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,

Pn = 'n (3).

EXERCISE 138

1. In how many orders may 5 persons sit on a bench? Substitute in the formula, P = '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

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