Exercise 155 Page 649

Practice makes perfect

a Do not get fooled by the word combinations. Since there can be no repeats, the order of selection matters. Therefore, what we are looking for is the number of permutations we have when selecting 3 numbers from a set of 22.

_(22)P_3=22!/(22-3)! ⇔ _(22)P_3=22!/19! Let's determine this on our graphing calculator.

There are 9240 locker combinations.

b No, the word is not appropriate since the order of selection matters. If a certain integer is chosen for the first digit, it cannot be selected as the second digit. For example, 357 and 537 are two different combinations.

c To determine the number of mathematical combinations, we have to calculate _(22)C_3.

_(22)C_3=_(22)P_3/3! ⇔ _(22)C_3=22!/19! 3! Let's determine this on our graphing calculator.

There are 1540 locker combinations. However, this does not make any sense because it implies that any combination that contain the integers of the combination unlocks the lock.

d Let's list some acceptable and unacceptable codes.

Acceptable:& 2 3 2 Unacceptable:& 2 3 3 or 2 2 3 When choosing the first digit, we have 22 numbers available. For the second digit, the first number is not available anymore which means there are 21 left we can choose. For the third digit, we have to exclude the second digit. However, the first digit is now available for a total of 21 numbers. With this information, we can calculate the number of choices. 22* 21* 21=9702