CHAPTER 1 Basic Probability

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(b) Denote the 3 labor representatives by L1, L2, L3; the management representatives by M1, M2; and the public representatives by P1, P2, P3, P4. Then the tree diagram of Fig. 1-10 shows that there are 24 different committees in all. From this tree diagram we can list all these different committees, e.g., L1 M1 P1, L1 M1 P2, etc.

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Fig. 1-10

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Permutations 1.19. In how many ways can 5 differently colored marbles be arranged in a row

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We must arrange the 5 marbles in 5 positions thus: . The first position can be occupied by any one of 5 marbles, i.e., there are 5 ways of filling the first position. When this has been done, there are 4 ways of filling the second position. Then there are 3 ways of filling the third position, 2 ways of filling the fourth position, and finally only 1 way of filling the last position. Therefore: Number of arrangements of 5 marbles in a row In general, Number of arrangements of n different objects in a row n(n l)(n 2) c 1 n! 5 4 3 2 l 5! 120

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This is also called the number of permutations of n different objects taken n at a time and is denoted by n Pn.

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1.20. In how many ways can 10 people be seated on a bench if only 4 seats are available

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The first seat can be filled in any one of 10 ways, and when this has been done, there are 9 ways of filling the second seat, 8 ways of filling the third seat, and 7 ways of filling the fourth seat. Therefore: Number of arrangements of 10 people taken 4 at a time In general, Number of arrangements of n different objects taken r at a time n(n 1) c (n r 1) 10 9 8 7 5040

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This is also called the number of permutations of n different objects taken r at a time and is denoted by nPr. Note that when r n, n Pn n! as in Problem 1.19.

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CHAPTER 1 Basic Probability

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1.21. Evaluate (a) 8 P3, (b) 6 P4, (c) l5 P1, (d) 3 P3.

(a) 8 P3 8 7 6 336 (b) 6P4 6 5 4 3 360 (c)

15 P1

15 (d) 3 P3

3 2 1

1.22. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible

The men may be seated in 5 P5 ways, and the women in 4 P4 ways. Each arrangement of the men may be associated with each arrangement of the women. Hence, Number of arrangements

5 P5

4 P4

5! 4!

(120)(24)

1.23. How many 4-digit numbers can be formed with the 10 digits 0, 1, 2, 3, . . . , 9 if (a) repetitions are allowed, (b) repetitions are not allowed, (c) the last digit must be zero and repetitions are not allowed

(a) The first digit can be any one of 9 (since 0 is not allowed). The second, third, and fourth digits can be any one of 10. Then 9 10 10 10 9000 numbers can be formed. (b) The first digit can be any one of 9 (any one but 0). The second digit can be any one of 9 (any but that used for the first digit). The third digit can be any one of 8 (any but those used for the first two digits). The fourth digit can be any one of 7 (any but those used for the first three digits). Then 9 9 8 7 4536 numbers can be formed.

Another method

The first digit can be any one of 9, and the remaining three can be chosen in 9 P3 ways. Then 9 9P3 9 9 8 7 4536 numbers can be formed. (c) The first digit can be chosen in 9 ways, the second in 8 ways, and the third in 7 ways. Then 9 8 7 numbers can be formed. 504