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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

2
3. 2 + V-3 and 2 -V-3.
4. 2 + 5V 3 and 2 – 5V3.
5. 1+v-1 and 1 - V-1.

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X =

Without solving, discuss the roots of the equation:

11. 3x2 + 17x 5 - 0.
12. 10.x2 21x 10 = 0.
13. 25x2 10x 1.
14. 3x2 5x + 4 0.
7 15

5
15.

+

= 0.
x + 2

8
16. 4x2 7x 10.
17. 2x2 X + 5 0.
18. 2x

1.
19. 7x2 + 13x 5.

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
ac bc 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 - (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

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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,

(3).

Pn

п

EXERCISE 138

1. In how
many orders

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

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