Probability: Mastering Permutations and Combinations Download PDF

Probability is the study of how likely or unlikely certain events are. Probability can help us make decisions, model phenomena, and understand uncertainty. To calculate probabilities, we often need to count the number of possible outcomes of an experiment or a situation. This is where permutations and combinations come in handy.

Permutations and combinations are two ways of arranging or selecting objects from a set. Permutations are arrangements where the order of the objects matters, while combinations are arrangements where the order does not matter. For example, if we have three letters A, B, and C, we can form six permutations: ABC, ACB, BAC, BCA, CAB, and CBA. However, we can form only one combination: ABC.

In this article, we will show you how to master permutations and combinations for probability problems. We will explain the formulas and rules for calculating permutations and combinations, and give you some examples and exercises to practice. We will also provide you with a PDF file that you can download for free and use as a reference or a study guide.

Permutations

A permutation is an ordered arrangement of r objects from a set of n objects. The notation for the number of permutations is nPr. For example, if we have 5 books and we want to arrange 3 of them on a shelf, the number of permutations is 5P3.

To calculate nPr, we can use the following formula:

nPr = n! / (n – r)!

In this formula, n! means n factorial, which is the product of all positive integers from 1 to n. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120. By convention, 0! = 1.

Using the formula, we can find that 5P3 = 5! / (5 – 3)! = 5! / 2! = (5 x 4 x 3 x 2 x 1) / (2 x 1) = 60.

This means that there are 60 ways to arrange 3 books out of 5 on a shelf.

Combinations

A combination is an unordered selection of r objects from a set of n objects. The notation for the number of combinations is nCr or (n choose r). For example, if we have 5 books and we want to choose 3 of them to read, the number of combinations is 5C3 or (5 choose 3).

To calculate nCr or (n choose r), we can use the following formula:

nCr = (n choose r) = n! / (r! x (n – r)!)

This formula is derived from the formula for permutations by dividing by r!, which is the number of ways to arrange r objects.

Using the formula, we can find that 5C3 = (5 choose 3) = 5! / (3! x (2!)) = (5 x 4 x 3 x 2 x 1) / ((3 x 2 x 1) x (2 x 1)) = 10.

This means that there are 10 ways to choose 3 books out of 5 to read.

Examples and Exercises

To help you master permutations and combinations for probability problems, here are some examples and exercises that you can try. You can also check your answers using the PDF file that we have provided at the end of this article.

Example 1

A password consists of four digits from 0 to 9. How many different passwords are possible?

Solution:

This is a permutation problem because the order of the digits matters. We have n = 10 digits and r = 4 positions. Using the formula for permutations, we get:

10P4 = 10! / (10 – 4)! = (10 x 9 x 8 x 7 x … ) / (6 x … ) = (10 x … ) / …

The answer is … different passwords.

Example …

Example 2

A committee of 5 people is to be formed from a group of 9 teachers and 12 students. How many different committees are possible?

Solution:

This is a combination problem because the order of the people in the committee does not matter. We have n = 9 + 12 = 21 people and r = 5 positions. Using the formula for combinations, we get:

21C5 = (21 choose 5) = 21! / (5! x (21 – 5)!) = (21 x 20 x 19 x 18 x 17 x … ) / ((5 x 4 x 3 x 2 x 1) x (16 x … )) = …

The answer is … different committees.

Example …

Example 3

A pizza shop offers 10 toppings to choose from. How many different pizzas with 4 toppings are possible?

Solution:

This is a combination problem because the order of the toppings does not matter. We have n = 10 toppings and r = 4 positions. Using the formula for combinations, we get:

10C4 = (10 choose 4) = 10! / (4! x (10 – 4)!) = (10 x 9 x 8 x 7 x … ) / ((4 x 3 x 2 x 1) x (6 x … )) = …

The answer is … different pizzas.

Exercise …

A lottery ticket consists of six numbers from 1 to 49. How many different lottery tickets are possible?

Solution:

This is a combination problem because the order of the numbers does not matter. We have n = 49 numbers and r = 6 positions. Using the formula for combinations, we get:

49C6 = (49 choose 6) = 49! / (6! x (49 – 6)!) = (49 x 48 x 47 x 46 x 45 x 44 x … ) / ((6 x 5 x 4 x 3 x 2 x 1) x (43 x … )) = …

The answer is … different lottery tickets.

Example 4

A restaurant offers 6 appetizers, 10 main courses, and 4 desserts. How many different meals consisting of one appetizer, one main course, and one dessert are possible?

Solution:

This is a permutation problem because the order of the dishes matters. We have n = 6 + 10 + 4 = 20 dishes and r = 3 positions. Using the formula for permutations, we get:

20P3 = 20! / (20 – 3)! = (20 x 19 x 18 x … ) / (17 x … ) = …

The answer is … different meals.

Exercise …

A license plate consists of three letters followed by three digits. How many different license plates are possible?

Solution:

This is a permutation problem because the order of the letters and digits matters. We have n = 26 letters and 10 digits and r = 6 positions. Using the formula for permutations, we get:

(26 + 10)P6 = 36P6 = 36! / (36 – 6)! = (36 x 35 x 34 x … ) / (30 x … ) = …

The answer is … different license plates.

Formulas for Permutations and Combinations

In this section, we will summarize the formulas for permutations and combinations that we have learned in the previous sections. We will also show how to use them to solve some probability problems.

The formula for the number of permutations of n objects taken r at a time is:

nPr = n! / (n – r)!

This means that we divide the factorial of n by the factorial of the difference between n and r.

The formula for the number of combinations of n objects taken r at a time is:

nCr = (n choose r) = n! / (r! x (n – r)!)

This means that we divide the factorial of n by the product of the factorials of r and the difference between n and r.

Examples and Exercises

To help you master the formulas for permutations and combinations, here are some examples and exercises that you can try. You can also check your answers using the PDF file that we have provided at the end of this article.

Example 5

A bag contains 5 red balls, 4 blue balls, and 3 green balls. How many ways are there to draw 3 balls from the bag?

Solution:

This is a combination problem because the order of the balls does not matter. We have n = 5 + 4 + 3 = 12 balls and r = 3 positions. Using the formula for combinations, we get:

12C3 = (12 choose 3) = 12! / (3! x (12 – 3)!) = (12 x 11 x 10 x … ) / ((3 x 2 x 1) x (9 x … )) = …

The answer is … ways.

Exercise …

A coin is tossed 5 times. How many ways are there to get exactly 3 heads?

Solution:

This is a combination problem because the order of the heads does not matter. We have n = 5 tosses and r = 3 heads. Using the formula for combinations, we get:

5C3 = (5 choose 3) = 5! / (3! x (5 – 3)!) = (5 x 4 x … ) / ((3 x … ) x (2 x … )) = …

The answer is … ways.

Quiz on Permutations and Combinations

To help you assess your understanding of permutations and combinations, here is a quiz that you can take. The quiz consists of 10 multiple-choice questions that cover the topics and concepts that we have learned in this article. You can also check your answers using the PDF file that we have provided at the end of this article.

Question 1

A coin is tossed 4 times. How many ways are there to get exactly 2 heads?

A) 4
B) 6
C) 8
D) 10

Question …

A bag contains 4 red balls, 5 blue balls, and 6 green balls. How many ways are there to draw 5 balls from the bag without replacement?

A) 3003
B) 252
C) 15
D) 120

Question 3

A group of 8 friends wants to take a photo. How many ways are there to arrange them in a row?

A) 8!
B) 8C8
C) 8P8
D) 8^8

Question …

A card is drawn at random from a standard deck of 52 cards. What is the probability that it is a king or a heart?

A) 4/52
B) 13/52
C) 16/52
D) 17/52

Question 5

A bag contains 4 red balls, 5 blue balls, and 6 green balls. How many ways are there to draw 5 balls from the bag without replacement?

A) 3003
B) 252
C) 15
D) 120

Question …

A license plate consists of three letters followed by three digits. How many different license plates are possible?

A) 17576
B) 15625
C) 1953125
D) 17576000

Solutions

The solutions to the quiz questions are as follows:

B) 6. This is a combination problem because the order of the heads does not matter. We have n = 4 tosses and r = 2 heads. Using the formula for combinations, we get: 4C2 = (4 choose 2) = 4! / (2! x (4 – 2)!) = (4 x … ) / ((2 x … ) x (2 x … )) = …
D) 17/52. This is a probability problem using the addition rule. The probability of drawing a king or a heart is equal to the probability of drawing a king plus the probability of drawing a heart minus the probability of drawing both a king and a heart. There are 4 kings and 13 hearts in a deck of 52 cards, but one of the hearts is also a king, so we have to subtract it from the total. Using fractions, we get: P(king or heart) = P(king) + P(heart) – P(king and heart) = (4/52) + (13/52) – (1/52) = (16/52) – (1/52) = …
A) 8!. This is a permutation problem because the order of the friends matters. We have n = 8 friends and r = 8 positions. Using the formula for permutations, we get: 8P8 = 8! / (8 – 8)! = …
A) 3003. This is a combination problem because the order of the balls does not matter. We have n = 4 + 5 + 6 = … balls and r = … positions. Using the formula for combinations, we get: … C … = (… choose …) = …! / (…! x (… – …)!) = (… x … x … x … ) / ((… x … x …) x (… x … )) = …
D) 17576000. This is a permutation problem because the order of the letters and digits matters. We have n = 26 letters and 10 digits and r = … positions. Using the formula for permutations, we get: (26 + …)P… = …P… = …! / (… – …)! = (… x … x … x …) / (… x …) = …

Conclusion

In this article, we have shown you how to master permutations and combinations for probability problems. We have explained the formulas and rules for calculating permutations and combinations, and given you some examples and exercises to practice. We have also provided you with a quiz to test your skills and a PDF file that you can download for free and use as a reference or a study guide.

Permutations and combinations are powerful tools that can help us count the number of possible outcomes of an experiment or a situation. By using permutations and combinations, we can solve complex probability problems that would otherwise be too tedious or impractical to do. Whether you are a student, a teacher, a researcher, or a curious learner, permutations and combinations can help you understand and appreciate the beauty and logic of probability.

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Probability: Mastering Permutations And Combinations Free Download Pdf 🖖🏿

Probability: Mastering Permutations and Combinations Download PDF

Permutations

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