lesson 6 problem solving practice permutations answer key

PERMUTATION PRACTICE PROBLEMS WITH ANSWERS

Problem 1 : 

Determine the number of permutations of the letters of the word SIMPLE if all are taken at a time?

Number of letters in the word "SIMPLE"  =  6

All are unique letters.

Number of permutation  = 6 P 6   =  6!

  =  6  ⋅ 5  ⋅ 4  ⋅ 3  ⋅ 2  ⋅ 1

  =  720

Hence total number of permutation is 720.

Problem 2 :

A test consists of 10 multiple choice questions. In how many ways can the test be answered if

(i) Each question has four choices?

(ii) The first four questions have three choices and the remaining have five choices?

(iii) Question number n has n + 1 choices?

Number of ways to answer 1 st question  =  4

Number of ways to answer 2 nd  question  =  4

Number of ways to answer 3 rd  question  =  4

............................

Number of ways  =  4  ⋅ 4  ⋅ 4  ⋅ 4  ⋅ 4  ⋅ 4  ⋅ 4  ⋅ 4  ⋅ 4  ⋅ 4

  =  4 10

Hence the total number of ways  =   4 10

Number of ways to answer 1 st question  =  3

Number of ways to answer 2 nd  question  =  3

Number of ways to answer 3 rd  question  =  3

Number of ways to answer 4 th  question  =  3

Number of ways to answer 5 th  question  =  5

Number of ways to answer 6 th  question  =  5

.............................. 

Number of ways  =    3  ⋅ 3  ⋅ 3  ⋅ 3  ⋅ 5  ⋅ 5   ⋅ 5  ⋅ 5  ⋅ 5  ⋅ 5

  =  3 4 ⋅ 5 6

Hence the total number of ways is  3 4 ⋅ 5 6 .

Number of ways to answer 1 st question  =  2

Number of ways to answer 2 nd question  =  3

Number of ways to answer 4 th  question  =  5

...................

Number of ways to answer 10 th  question  =  11

Number of ways  =  1  ⋅  2  ⋅  3  ⋅  4  ⋅  5  ⋅  6  ⋅  7  ⋅  8  ⋅ 9  ⋅  10  ⋅  11

  =  11!

Hence the total number of ways is 11!.

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Free Printable permutations worksheets

Discover a vast collection of free printable Math permutations worksheets, carefully crafted to help students enhance their understanding of this essential topic. Ideal for Math teachers and learners alike.

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Explore printable permutations worksheets

Permutations worksheets are an essential tool for teachers who want to help their students master the art of solving problems related to permutations in Math and Algebra. These worksheets provide a variety of exercises and problems that challenge students to apply their knowledge of permutations, combinations, and other related concepts. Teachers can use these worksheets to supplement their lesson plans and provide additional practice opportunities for their students. By incorporating permutations worksheets into their curriculum, educators can ensure that their students develop a strong foundation in Math and Algebra, enabling them to tackle more complex problems with confidence.

Quizizz is an innovative platform that offers a wide range of educational resources, including permutations worksheets, to support teachers in their quest to provide engaging and effective learning experiences for their students. In addition to worksheets, Quizizz also offers interactive quizzes, games, and other learning tools that can be easily integrated into lesson plans. This platform allows teachers to create customized quizzes and assignments, track student progress, and provide real-time feedback to help students improve their understanding of Math and Algebra concepts. By leveraging the power of Quizizz, educators can enhance their teaching methods and ensure that their students are well-prepared to excel in their studies.

Permutations Worksheets

What Are Statistical Permutations? Probability is the mathematical calculation of the chances of an event taking place. It is the methodology that is used by weather forecasting professionals and even in lotteries. Therefore, learning probability might also help you win a lottery! So, there are two types of probabilities when it comes to finding the number of arrangements of objects. These include permutations and combinations. These two can confuse the students, and it is what makes it seem complicated. A permutation is an arrangement of objects together while following a fixed order. When we are talking about arrangement without order, we call this type of probability, combinations. To calculate permutations, you need to use this formula; nPr = n!/(n-r)! Here n is the number of elements in the set that need to permuted, r is size of each permutation. Both of these are non-negative integers.

Basic Lesson

Introduces the fundamentals of Permutations. Provides a basic application.

Intermediate Lesson

This lesson focuses on determining the outcome of Permutations with word problems.

Independent Practice 1

Students practice with 20 Permutations problems. The answers can be found below.

Independent Practice 2

Another 20 Permutations problems. The answers can be found below.

Homework Worksheet

Reviews all skills in the unit. A great take home sheet. Also provides a practice problem.

10 problems that test Permutations skills.

Homework and Quiz Answer Key

Answers for the homework and quiz.

Answers for the lesson and practice sheets.

Make Believe...

A new study has shown that 64.386429% of all statistics are made up on the spot. Did you know that 74.7116932% of all statistics claim a precision of results not able to be proven by the method used?

• Maths Questions

Permutation Questions

Permutation questions deal with the arrangement of objects in a specific order or formation of a number of different words from the letters of a given word, etc. There exist a variety of cases in which we apply permutations to get the possible results. In this article, you will get solved questions on permutation, and some practice questions for the same.

What is Permutation?

Permutation refers to the arrangement of objects in a definite order. That means permutation is the arrangement of objects in which order matters. The arrangement of r objects out of n objects can be calculated using the permutation formula. That is:

n P r = n!/(n – r)!

Learn in detail about permutation here.

Permutation Questions and Answers

1. Calculate the following:

(i) n P r when n = 12, r = 5

n P r = 12 P 5 = 12!/(12 – 5)!

= (12 × 11 × 10 × 9 × 8 × 7!)/7!

= 12 × 11 × 10 × 9 × 8

9 P 4 = 9!/(9 – 4)! = 9!/5!

= (9 × 8 × 7 × 6 × 5!)/5!

2. In how many different ways can the letters of the word THOUGHTS be arranged so that the vowels always come together?

Given word: THOUGHTS

Number of letters = 8

T’s = 2

H’s = 2

Number of vowels = 2 (O, U)

Vowels should come together.

So, the number of letters for arrangement = 7

i.e., (OU)THGHTS

Number of arrangements = 7!

And two vowels can be arranged in 2! ways.

Therefore, the total number of ways of arrangements = (7! × 2!)/(2! 2!)

= (7 × 6 × 5 × 4 × 3 × 2!)/2!

3. In how many ways can seven books be arranged on a shelf?

Number of ways in which the first book can be placed = 7

Number of ways in which the second book can be placed = 6

The total number of ways in which seven books can be arranged on a shelf = 7 × 6 × 5 × 4 × 3 × 2 × 1 (i.e., 7!)

4. How many different arrangements of letters of the word MATHEMATICS are possible?

Given word: MATHEMATICS

Number of letters = 11

Number of different arrangements = 11!/(2! 2! 2!)

= (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(2 × 1 × 2 × 1 × 2 × 1)

5. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, and 7 if no digit is repeated?

Given digits: 1, 2, 3, 4, 6, 7

Number of digits = 6

Number of possible digits at unit’s place = 3 (2, 4 and 6)

⇒ Number of permutations = 3 P 1 = 3

When one of the digits is taken in units’ place, then the number of possible digits available = 5

⇒ Number of permutations = 5 P 2 = 5!/(5 – 2)! = 5!/3! = 120/6 = 20

The total number of permutations = 3 × 20 = 60.

Therefore, 60 three-digit numbers can be made using the given digits.

6. 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?

Given: 5 men and 4 women

Total number of people = 9

The women occupy even places, which means they will be sitting in the 2nd, 4th, 6th and 8th places, whereas the men will be sitting in the 1st, 3rd, 5th, 7th and 9th places.

The number of arrangements in which 4 women can sit in 4 places = 4 P 4 = 4!/(4 – 4)! = 4!/0! = 24/1 = 24

5 men can occupy 5 seats in 5 ways.

That means the number of ways they can be seated = 5 P 5 = 5!/(5 – 5)! = 5!/0! = 120/1 = 120

Therefore, the total numbers of possible sitting arrangements = 24 × 120 = 2880

7. How many numbers are there between 100 and 1000 such that at least one of their digits is 7?

Total number of 3-digit numbers having at least one of their digits as 7 = (Total number of 3-digit numbers) – (Total number of 3-digit numbers in which 7 does not appear at all)

Let us find the total number of 3-digit numbers between 100 and 1000.

That means repetition of digits is allowed.

The hundred’s place can be filled in 9 ways, i.e., using digits from 1 to 9.

The ten’s and the unit’s place can be filled in 10 ways using the digits from 0 to 9.

∴ Total number of 3-digit numbers = 9 × 10 × 10 = 900

Now, we need to find the total number of 3-digit numbers in which 7 does not appear.

Here, 9 digits to be used, i.e., 0, 1, 2, 3, 4, 5, 6, 8, 9.

Now, the hundred’s place can be filled in 8 ways (excluding 0), and the tens’ and ones’ place can be filled in 9 ways each.

∴ Total number of 3-digit numbers in which 7 does not appear = 8 × 9 × 9 = 648

Hence, the numbers between 100 and 1000 in which at least one of their digits is 7 = 900 – 648 = 252

8. P, Q, R, S, and T sit on five chairs facing north. R will sit only on the leftmost chair, and Q will not sit anywhere to the left of P. In how many ways can they be seated?

R will sit on 1 and Q will sit somewhere to the right of P

R will sit on 1

Then there are three possible ways.

P on 2, so Q can be seated on 3, 4 or 5

The remaining two can be seated on two chairs in 2 ways

Number of possible ways = 3 × 2 = 6

P on 3, so Q can be seated on 4 or 5

Number of possible ways = 2 × 2 = 4

P on 4, so Q will be on 5

Number of possible ways = 2

Total number of possible ways = (6 + 4 + 2) = 12

9. In how many ways can 10 differently coloured beads be threaded on a string?

As the necklace can be turned over, clockwise and anti-clockwise arrangements are the same.

Also, the number of string arrangements of n objects = (n – 1)!/2

Therefore, the number of ways in which 10 differently coloured beads can be threaded on a string = (10 – 1)!/2 = 9!/2

10. How many 3 digit numbers can be formed with the digits 5, 6, 2, 3, 7 and 9, which are divisible by 5, and none of its digits are repeated?

If the number has unit’s digits as 0 or 5, then it will be divisible by 5.

Given digits: 5, 6, 2, 3, 7, 9

The digit that can be placed at the unit’s place so that the three-digit number is divisible by 5 is 5.

This can be done in 1 way.

Now, the tens and hundreds of places can be filled in 5 and 4 ways since the repetition of the digits are not allowed.

Therefore, the total number of such three digits numbers = 5 × 4 × 1 = 20

Practice Problems on Permutation

• How many ways can the letters of the word “EXAMINATION” be arranged such that the first and last letters are the same, and the vowels are together?
• How many numbers between 100 and 1000 use only odd digits, no digit being repeated?
• How many 3-letter words, with or without meaning, can be formed out of the letters of the word LOGARITHMS, if repetition of letters is not allowed?
• In how many ways can 9 different colour balls be arranged in a row so that the black, white, red and green balls are never together?
• Find the sum of all the possible numbers of 4 digits formed by digits 3, 5, 5, and 6 using each digit once.

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Curriculum  /  Math  /  6th Grade  /  Unit 6: Equations and Inequalities  /  Lesson 6

Equations and Inequalities

Lesson 6 of 14

Criteria for Success

Tips for teachers, anchor problems, problem set, target task, additional practice.

Solve percent problems using equations.

Common Core Standards

Core standards.

The core standards covered in this lesson

Expressions and Equations

6.EE.B.7 — Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q and x are all nonnegative rational numbers.

Ratios and Proportional Relationships

6.RP.A.3.C — Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.

Foundational Standards

The foundational standards covered in this lesson

6.RP.A.1 — Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, "The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak." "For every vote candidate A received, candidate C received nearly three votes."

6.RP.A.2 — Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. Expectations for unit rates in this grade are limited to non-complex fractions. For example, "This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar." "We paid $75 for 15 hamburgers, which is a rate of $5 per hamburger."

The essential concepts students need to demonstrate or understand to achieve the lesson objective

• Determine, through repeated reasoning, an equation to represent the relationship between percent, whole, and part:   $$percent\times{whole}=part$$   (MP.8).
• Write an equation to represent a percent situation when given a part and a percent.
• Write and solve equations to find the whole, given the part and percent.

Suggestions for teachers to help them teach this lesson

In Unit 2, students solved percent problems by reasoning about diagrams, double number lines, and tables. Now having learned about equations in the form  $${px=q}$$ , students revisit percent problems to see how they can be modeled and solved efficiently using an equation (MP.4).

Unlock features to optimize your prep time, plan engaging lessons, and monitor student progress.

Problems designed to teach key points of the lesson and guiding questions to help draw out student understanding

25-30 minutes

• 30% of 120?

In general, if you’re given a percent and a whole, how can you find the part? Write this as an equation.

Guiding Questions

30% of what number is 12?

Solve this problem first by drawing a diagram. Then write and solve an equation to verify your solution. 

For each situation below, write and solve an equation to answer the question.

a.   There are 6 liters of water in a bucket, which is 20% of the maximum number of liters the bucket can hold. What is the maximum number of liters the bucket can hold?

b.   A softball team won 18 games, which was 60% of the games they played this season. How many games did the softball team play this season?

A set of suggested resources or problem types that teachers can turn into a problem set

15-20 minutes

Give your students more opportunities to practice the skills in this lesson with a downloadable problem set aligned to the daily objective.

A task that represents the peak thinking of the lesson - mastery will indicate whether or not objective was achieved

5-10 minutes

Paula is saving money to buy a tablet. So far, she has saved $54, which is 45% of what she needs to buy the tablet. 

Write and solve an equation to find the price of the tablet. 

Student Response

An example response to the Target Task at the level of detail expected of the students.

The following resources include problems and activities aligned to the objective of the lesson that can be used for additional practice or to create your own problem set.

• Challenge: In a choir, there are 28 female singers which is 40% of the choir. How many female singers would have to be added to the group so exactly 50% of the choir were females?
• EngageNY Mathematics Grade 6 Mathematics > Module 1 — Lessons 27–29 (Revisit these lessons from Unit 2 and have students write equations to solve.)
• Open Up Resources Grade 6 Unit 6 Practice Problems — Lesson 7 #1–3

Topic A: Reasoning About and Solving Equations

Represent equations in the form  $${ x+p=q }$$ and  $${px=q}$$ using tape diagrams and balances.

6.EE.B.6 6.EE.B.7

Define and identify solutions to equations.

Write equations for real-world situations.

Solve one-step equations with addition and subtraction.

Solve one-step equations with multiplication and division.

6.EE.B.7 6.RP.A.3.C

Solve multi-part equations leading to the form  $${x+p=q }$$  and $${px=q}$$ .

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Topic B: Reasoning About and Solving Inequalities

Define and identify solutions to inequalities.

6.EE.B.5 6.EE.B.8

Write and graph inequalities for real-world conditions. (Part 1)

Write and graph inequalities for real-world conditions. (Part 2)

Solve one-step inequalities.

6.EE.B.6 6.EE.B.8

Topic C: Representing and Analyzing Quantitative Relationships

Write equations for and graph ratio situations. Define independent and dependent variables.

6.EE.C.9 6.RP.A.3.A

Represent the relationship between two quantities in graphs, equations, and tables. (Part 1)

Represent the relationship between two quantities in graphs, equations, and tables. (Part 2)

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