answer: c(47,5)xc(27,1) = 1,533,939 x 27 = 41,416,353

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In SuperLotto Plus, a California state lottery game, you select five distinct numbers from 1 to 47, and one MEGA number from 1 to 27, hoping that your selection will match a random list selected by lottery officials. (a) How many different sets of six numbers can you select?

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In SuperLotto Plus, a California state lottery game, you select five distinct numbers from 1 to 47, and one MEGA number from 1 to 27, hoping that your selection will match a random list selected by lottery officials. (a) How many different sets of six numbers can you select?. - PowerPoint PPT Presentation

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Page 1: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

In SuperLotto Plus, a California state lottery game, you select five distinct numbers from 1 to 47, and one MEGA number from 1 to 27, hoping that your selection will match a random list selected by lottery officials.

(a) How many different sets of six numbers can you select?

Page 2: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

In SuperLotto Plus, a California state lottery game, you select five distinct numbers from 1 to 47, and one MEGA number from 1 to 27, hoping that your selection will match a random list selected by lottery officials.

(a) How many different sets of six numbers can you select?

Answer: C(47,5)xC(27,1)

= 1,533,939 x 27

= 41,416,353

Page 3: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

In SuperLotto Plus, a California state lottery game, you select five distinct numbers from 1 to 47, and one MEGA number from 1 to 27, hoping that your selection will match a random list selected by lottery officials.

Eileen Burke always includes her age and her husband’s age as two of the first five numbers in her SuperLotto Plus selections. How many ways can she complete her list of six numbers?

Page 4: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

In SuperLotto Plus, a California state lottery game, you select five distinct numbers from 1 to 47, and one MEGA number from 1 to 27, hoping that your selection will match a random list selected by lottery officials.

Eileen Burke always includes her age and her husband’s age as two of the first five numbers in her SuperLotto Plus selections. How many ways can she complete her list of six numbers?

Answer: C(45,3)xC(27,1)

= 14,190x 27

= 383,130

Page 5: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Drawing Cards - How many cards must be drawn (without replacement) from a standard deck of 52 to guarantee drawing two cards of the same suit?

Page 6: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Drawing Cards - How many cards must be drawn (without replacement) from a standard deck of 52 to guarantee drawing two cards of the same suit?

Answer: 5

Page 7: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Drawing Cards - How many cards must be drawn (without replacement) from a standard deck of 52 to guarantee drawing three cards of the same suit?

Page 8: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Drawing Cards - How many cards must be drawn (without replacement) from a standard deck of 52 to guarantee drawing three cards of the same suit?

Answer: 9

Page 9: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many different 5-card poker hands would contain only cards of a single suit?

Page 10: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many different 5-card poker hands would contain only cards of a single suit?

Answer: 4 x C(13,5)

= 5148

Page 11: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Subject identification numbers in a certain scientific research project consist of three letters followed by three digits and then three more letters. Assume repetitions are not allowed within any of the three groups, but letters in the first group of three may occur also in the last group of three. How many distinct identification numbers are possible?

Page 12: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Subject identification numbers in a certain scientific research project consist of three letters followed by three digits and then three more letters. Assume repetitions are not allowed within any of the three groups, but letters in the first group of three may occur also in the last group of three. How many distinct identification numbers are possible?

Answer: P(26,3) x P(10,3) x P(26,3)

= 175,219,200,000

Page 13: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Radio stations in the United States have call letters that begin with K or W (for west or east of the Mississippi River, respectively). Some have three call letters, such as WBZ in Boston, WLS in Chicago, and KGO in San Francisco. Assuming no repetition of letters, how many three-letter sets of call letters are possible?

Page 14: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Radio stations in the United States have call letters that begin with K or W (for west or east of the Mississippi River, respectively). Some have three call letters, such as WBZ in Boston, WLS in Chicago, and KGO in San Francisco. Assuming no repetition of letters, how many three-letter sets of call letters are possible?

Answer: 2 x P(25,2)= 1200

Page 15: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Most stations that were licensed after 1927 have four call letters starting with K or W, such as WXYZ in Detroit or KRLD in Dallas. Assuming no repetitions, how many four-letter sets are possible?

Page 16: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Most stations that were licensed after 1927 have four call letters starting with K or W, such as WXYZ in Detroit or KRLD in Dallas. Assuming no repetitions, how many four-letter sets are possible?

Answer: 2 x P(25,3)= 27,600

Page 17: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Each team in an eight-team basketball league is scheduled to play each other team three times. How many games will be played altogether?

Page 18: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Each team in an eight-team basketball league is scheduled to play each other team three times. How many games will be played altogether?

Answer: 3 x C(8,2)= 84

Page 19: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

The Coyotes, a youth league baseball team, have seven pitchers, who only pitch, and twelve other players, all of whom can play any position other than pitcher. For Saturday’s game, the coach has not yet determined which nine players to use nor what the batting order will be, except that the pitcher will bat last. How many different batting orders may occur?

Page 20: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

The Coyotes, a youth league baseball team, have seven pitchers, who only pitch, and twelve other players, all of whom can play any position other than pitcher. For Saturday’s game, the coach has not yet determined which nine players to use nor what the batting order will be, except that the pitcher will bat last. How many different batting orders may occur?

Answer: 7 x P(12,8)= 139,708,800

Page 21: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A music class of eight girls and seven boys is having a recital. If each member is to perform once, how many ways can the program be arranged so that a girl always performs first?

Page 22: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A music class of eight girls and seven boys is having a recital. If each member is to perform once, how many ways can the program be arranged so that a girl always performs first?

Answer: 8 x 14!= 697,426,329,600

Page 23: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A music class of eight girls and seven boys is having a recital. If each member is to perform once, how many ways can the program be arranged so that a girl always performs first and a boy always performs second?

Page 24: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A music class of eight girls and seven boys is having a recital. If each member is to perform once, how many ways can the program be arranged so that a girl always performs first and a boy always performs second?

Answer: 8 x 7 x 13!= 348,713,164,800

Page 25: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A music class of eight girls and seven boys is having a recital. If each member is to perform once, how many ways can the program be arranged so that Lisa always performs first and Doug always performs second?

Page 26: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A music class of eight girls and seven boys is having a recital. If each member is to perform once, how many ways can the program be arranged so that Lisa always performs first and Doug always performs second?

Answer: 1x13!x1= 6,227,020,800

Page 27: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A music class of eight girls and seven boys is having a recital. If each member is to perform once, how many ways can the program be arranged so that the entire program will alternate between girls and boys?

Page 28: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A music class of eight girls and seven boys is having a recital. If each member is to perform once, how many ways can the program be arranged so that the entire program will alternate between girls and boys?

Answer: 8! x 7!= 203,212,800

Page 29: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A music class of eight girls and seven boys is having a recital. If each member is to perform once, how many ways can the program be arranged so that the first, eighth and fifteenth performers must be girls?

Page 30: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A music class of eight girls and seven boys is having a recital. If each member is to perform once, how many ways can the program be arranged so that the first, eighth and fifteenth performers must be girls?

Answer: 8 x 7 x 6 x 12!= 160,944,537,600

Page 31: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Carole begins each day by reading from one of seven inspirational books. How many ways can she choose the books for one week if the selection is done by placing the book back on the bookshelf after she reads it?

Page 32: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Carole begins each day by reading from one of seven inspirational books. How many ways can she choose the books for one week if the selection is done by placing the book back on the bookshelf after she reads it?

Answer: 77

= 823,543

Page 33: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Carole begins each day by reading from one of seven inspirational books. How many ways can she choose the books for one week if the selection is done by choosing a different book each day?

Page 34: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Carole begins each day by reading from one of seven inspirational books. How many ways can she choose the books for one week if the selection is done by choosing a different book each day?

Answer: 7!= 5,040

Page 35: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many of the possible 5-card hands from a standard 52-card deck would consist of four clubs and one non-club?

Page 36: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many of the possible 5-card hands from a standard 52-card deck would consist of four clubs and one non-club?

Answer: C(13,4) x 39

= 27,885

Page 37: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many of the possible 5-card hands from a standard 52-card deck would consist of two face cards and three non-face cards?

Page 38: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many of the possible 5-card hands from a standard 52-card deck would consist of two face cards and three non-face cards?

Answer: C(12, 2) x C(40, 3)

= 652, 080

Page 39: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many of the possible 5-card hands from a standard 52-card deck would consist of two red cards two clubs and a spade?

Page 40: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many of the possible 5-card hands from a standard 52-card deck would consist of two red cards two clubs and a spade?

Answer: C(26, 2) x C(13, 2) x 13

= 329,550

Page 41: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

In how many ways could twenty-five people be divided into five groups containing, respectively, three, four, five, six, and seven people?

Page 42: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

In how many ways could twenty-five people be divided into five groups containing, respectively, three, four, five, six, and seven people?

Answer: C(25,3) x C(22,4) x C(18,5) x C(13,6)

= 2.474 x 1014

Page 43: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many different three-number “combinations” are possible on a combination lock having 40 numbers on its dial? (Hint: “Combination” is a misleading name for these locks since repetitions are allowed and also order makes a difference.)

Page 44: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many different three-number “combinations” are possible on a combination lock having 40 numbers on its dial? (Hint: “Combination” is a misleading name for these locks since repetitions are allowed and also order makes a difference.)

Answer : 403

= 64,000

Page 45: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Michael Grant, his wife and son, and four additional friends are driving, in two vehicles, to the seashore. If all seven people are available to drive, how many ways can the two drivers be selected? (Everyone would like to drive the sports car, so it is important which driver gets which car.)

Page 46: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Michael Grant, his wife and son, and four additional friends are driving, in two vehicles, to the seashore. If all seven people are available to drive, how many ways can the two drivers be selected? (Everyone would like to drive the sports car, so it is important which driver gets which car.)

Answer: P(7,2)

= 210

Page 47: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

At the race track, you win the “daily double” by purchasing a ticket and selecting the winners of both of two specified races. If there are six and eight horses running in the first and second races, respectively, how many tickets must you purchase to guarantee a winning selection?

Page 48: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

At the race track, you win the “daily double” by purchasing a ticket and selecting the winners of both of two specified races. If there are six and eight horses running in the first and second races, respectively, how many tickets must you purchase to guarantee a winning selection?

Answer: 6 x 8

= 48

Page 49: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Many race tracks offer a “trifecta” race. You win by selecting the correct first-, second-, and third-place finishers. If eight horses are entered, how many tickets must you purchase to guarantee that one of them will be a trifecta winner?

Page 50: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Many race tracks offer a “trifecta” race. You win by selecting the correct first-, second-, and third-place finishers. If eight horses are entered, how many tickets must you purchase to guarantee that one of them will be a trifecta winner?

Answer: P(8,3)

= 336

Page 51: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Because of his good work, Jeff Hubbard gets a contract to build homes on three additional blocks in the subdivision, with six homes on each block. He decides to build nine deluxe homes on these three blocks: two on the first block, three on the second, and four on the third. The remaining nine homes will be standard. Altogether on the three-block stretch, how many different choices does Jeff have for positioning the eighteen homes? (Hint: Consider the three blocks separately and use the fundamental counting principle.)

Page 52: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

Because of his good work, Jeff Hubbard gets a contract to build homes on three additional blocks in the subdivision, with six homes on each block. He decides to build nine deluxe homes on these three blocks: two on the first block, three on the second, and four on the third. The remaining nine homes will be standard. Altogether on the three-block stretch, how many different choices does Jeff have for positioning the eighteen homes? (Hint: Consider the three blocks separately and use the fundamental counting principle.)

Answer: C(6,2) x C(6,3) x C(6,4)

= 4500

Page 53: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many six-digit counting numbers can be formed using all six digits 4, 5, 6, 7, 8, and 9?

Page 54: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many six-digit counting numbers can be formed using all six digits 4, 5, 6, 7, 8, and 9?

Answer: 6! or P(6,6)

= 720

Page 55: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A professor teaches a class of 60 students and another class of 40 students. Five percent of the students in each class are to receive a grade of A. How many different ways can the A grades be distributed?

Page 56: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A professor teaches a class of 60 students and another class of 40 students. Five percent of the students in each class are to receive a grade of A. How many different ways can the A grades be distributed?

Answer: C(60,3) x C(40,2)

= 26,691,600

Page 57: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many counting numbers have four distinct nonzero digits such that the sum of the four digits is 12?

Page 58: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

How many counting numbers have four distinct nonzero digits such that the sum of the four digits is 12?

Answer: 2 x 4!

= 48

Only 2 ways to get a sum of 12

Using 1,2,3 and 6 or 1,2,4 and 5

Page 59: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A computer company will screen a shipment of 30 processors by testing a random sample of five of them. How many different samples are possible?

Page 60: Answer: C(47,5)xC(27,1) = 1,533,939 x 27 = 41,416,353

A computer company will screen a shipment of 30 processors by testing a random sample of five of them. How many different samples are possible?

Answer: C(30,5)

= 142,506