Section 4.7 Solutions

1) The daily double at most racetracks consists of selecting the winning horse in both the first and the second race. If the first race has 7 entries and the second has 8 entries, how many daily double tickets mush you purchase to guarantee a win? __7__*___8__ = 56 1st race 2nd race Answer: 56 3) A lock combination consists of three numbers between 0 and 24 inclusive. How many combinations are possible, assume repetition is allowed? __25__*___25__*__25__ = 15,625 1st # 2nd 3rd Answer: 15,625 5) A social security number consists of nine digits. How many different social security numbers are possible if repetition of digits is NOT permitted? ___10___*____9____*___8___ *___7___*___6___*___5___*___4___*___3___*___2___ 1st # 2nd 3rd 4th 5th 6th 7th 8th 9th Answer: 3,628,800 7) A class ring has three stone options (emerald, ruby or sapphire), two metal options (gold and silver). How many different class rings are possible? __10__*__2__ = 20 Stone metal Answer: 20 9) How many different sandwiches can be made with 2 choices of bread, 3 choices of toppings and 2 choices of meat? __2__*__3____*__2__ = 12 Bread toppings meat Answer: 12 11) A batting order must be created from the 9 starters on a baseball team. How many batting orders are possible if the pitcher must bat last? First place the pitcher. _______*________*________ *______*______*______*______*______*___1__ 1st # 2nd 3rd 4th 5th 6th 7th 8th 9th Now there are only 8 choices for the 1st hitter. Increment down as each person can only bat once. ___8___*____7____*___6___ *___5___*___4___*___3___*___2___*___1___*___1__ 1st # 2nd 3rd 4th 5th 6th 7th 8th 9th Answer: 40,320 13) A batting order must be created from the 9 starters on a baseball team. How many batting orders are possible if the first 4 batters are already assigned? Place the first 4 hitters. ___1____*____1____*___1_____ *___1___*______*______*______*______*_______ 1st # 2nd 3rd 4th 5th 6th 7th 8th 9th Now there are only 5 choices for the 5th hitter. Increment down as each person can only bat once. ___1___*____1____*___1___ *___1___*___5___*___4___*___3___*___2___*___1__ 1st # 2nd 3rd 4th 5th 6th 7th 8th 9th Answer: 120 15) A multiple choice test has 5 questions and each question has 4 choices. How many ways are there to answer the questions if you know the answer to the first question? There is only one choice for my first answer. Each additional question has 4 possible guesses. __1___*___4___*___4___*___4___*___4___ 1st 2nd 3rd 4th 5th Answer: 256 Homework 17 – 20: an identification code is to consist of two letters followed by 4 digits. How many different codes are possible if 17) Repetition is permitted ___26___*___26___*___10___*___10___*___10___*___10___ = 6,760,000 L L # # # # Answer: 6,760,000 19) Repetition of numbers is permitted, but repetition of letters is not permitted? ___26___*___25___*___10___*___10___*___10___*___10___ = 6,760,000 L L # # # # Answer: 6,500,000 Homework #21-24: a license plate is to consist of three digits followed by two uppercase letters. Determine the number of different license plates possible if 21) Repetition of numbers and letters is not permitted ___10___*___9___*___8___*___26___*___25___ = 468,000 # # # L L Answer: 468,000 23) The first digit cannot be a zero, and repetition is not permitted The first digit must be any of 9 numbers other than zero. I will write 9 first. Next there are 9 numbers that have not been placed (1,2,3,4,5,6,7,8,9) the next number will be 9. I will increment down for the rest of the numbers. ___9___*___9___*___8___*___26___*___25___ = 421,200 # # # L L Answer: 441,200 Homework #25-30: Use the permutation of distinct objects formula n! to solve the n1 ! n2 ! n3 ! nr ! following 25) How many ways can the letters of Mississippi be arranged? There are 11 letters. So numerator is 11! There are 4 i’s, 4 s’s, 2 p’s. The denominator will be 4!*4!*2! 11! 11 ∗ 10 ∗ 9 ∗ 8 ∗ 7 ∗ 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1 = = 34,650 4! ∗ 4! ∗ 2! 4∗3∗2∗1∗4∗3∗2∗1∗2∗1 Answer: 34,650 27) How many ways can the letters x,x,x,y,y,z,z,z,z be arranged? There are 10 letters. The numerator will be 10! The denominator will be 3!*2!*4! as there are 3 x’s, 2 y’s and 4 z’s. 9! 9∗8∗7∗6∗5∗4∗3∗2∗1 = = 1260 3! ∗ 2! ∗ 4! 3 ∗ 2 ∗ 1 ∗ 2 ∗ 1 ∗ 4 ∗ 3 ∗ 2 ∗ 1 Answer: 1,260 ways 29) How many ways can the letters of abracadabra be arranged? 11 letters, 5 a’s, 2 b’s, 2 r’s 11! = 5!∗2!∗2! 11∗10∗9∗8∗7∗6∗5∗4∗3∗2∗1 5∗4∗3∗2∗1∗2∗1∗2∗1 Answer: 83,160 ways = 83,160