Problems in Permutations

Here are more explaination about Permutations. Examples and solutions for you :)

Definition

" Permutations are the different ways in which a collection of items can be arranged " 

For example:

The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are ABC, ACB, BCA, CBA, CAB, BAC.

Note that ABC and CBA are not same as the order of arrangement is different. The same rule applies while solving any problem in Permutations.
The number of ways in which n things can be arranged, taken all at a time, n P n = n!, called ‘n factorial.’

Factorial Formula:

Factorial of a number n is defined as the product of all the numbers from n to 1.

For example, the factorial of 5, 5! = 5*4*3*2*1 = 120.
Therefore, the number of ways in which the 3 letters can be arranged, taken all a time, is 3! = 3*2*1 = 6 ways.
Number of permutations of n things, taken r at a time, denoted by:

n P r = n! / (n-r)!

For example:

The different ways in which the 3 letters, taken 2 at a time, can be arranged is 3!/(3-2)! = 3!/1! = 6 ways.

Important Permutation Formulas:

1! = 1
0! = 1

Let us take a look at some examples:

Problem 1: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘CHAIR’.

Solution :

‘CHAIR’ contains 5 letters.

Therefore, the number of words that can be formed with these 5 letters = 5! = 5*4*3*2*1 = 120.
Problem 2: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘INDIA’.

Solution :

The word ‘INDIA’ contains 5 letters and ‘I’ comes twice.
When a letter occurs more than once in a word, we divide the factorial of the number of all letters in the word by the number of occurrences of each letter.
Therefore, the number of words formed by ‘INDIA’ = 5!/2! = 60.

Problem 3: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘SWIMMING?

Solution :

The word ‘SWIMMING contains 8 letters. Of which, I occurs twice and M occurs twice.
Therefore, the number of words formed by this word = 8! / (2!*2!) = 10080.

Problem 4: How many different words can be formed with the letters of the word ‘SUPER’ such that the vowels always come together?

Solution :

The word ‘SUPER’ contains 5 letters.
In order to find the number of permutations that can be formed where the two vowels U and E come together.

In these cases, we group the letters that should come together and consider that group as one letter.
So, the letters are S,P,R, (UE). Now the number of words are 4.

Therefore, the number of ways in which 4 letters can be arranged is 4!

In U and E, the number of ways in which U and E can be arranged is 2!

Hence, the total number of ways in which the letters of the ‘SUPER’ can be arranged such that vowels are always together are 4! * 2! = 48 ways.

Problem 5: Find the number of different words that can be formed with the letters of the word ‘BUTTER’ so that the vowels are always together.

Solution :

The word ‘BUTTER’ contains 6 letters.
The letters U and E should always come together. So the letters are B, T, T, R, (UE).
Number of ways in which the letters above can be arranged = 5!/2! = 60 (since the letter ‘T’ is repeated twice).

Number of ways in which U and E can be arranged = 2! = 2 ways

Therefore, total number of permutations possible = 60*2 = 120 ways.

Problem 6: Find the number of permutations of the letters of the word ‘REMAINS’ such that the vowels always occur in odd places.

Solution :

The word ‘REMAINS’ has 7 letters.
There are 4 consonants and 3 vowels in it.
Writing in the following way makes it easier to solve these type of questions.
(1) (2) (3) (4) (5) (6) (7)

No. of ways 3 vowels can occur in 4 different places = 4 P 3 = 24 ways.

After 3 vowels take 3 places, no. of ways 4 consonants can take 4 places = 4 P 4 = 4! = 24 ways.
Therefore, total number of permutations possible = 24*24 = 576 ways.

** More or less hoping that I can help you to understand more about PERMUTATION.

Problems in Combinations

Here are more explainations about Combinations. Problems and solutions for you :)

Definition

" The different selections possible from A collection of items are called combinations " 

For example:

The different selections possible from the alphabets A, B, C, taken 2 at a time, are AB, BC and CA.
It does not matter whether we select A after B or B after A. The order of selection is not important in combinations.

To find the number of combinations possible from a given group of items n, taken r at a time, the formula, denoted by:

n C r is
n C r = n! / [r! * (n-r)!]

For example, verifying the above example, the different selections possible from the alphabets A, B, C, taken two at a time are:

3 C2 = 3! / (2! * (3-2)!) = 3 possible selections (i.e., AB, BC, CA)

Important Combination formula:

n C n = 1
n C 0 = 1
n C 1 = n
n C r = n C(n-r)

The number of selections possible with A, B, C, taken all at a time is 3C 3 = 1 (i.e. ABC)

Solved examples of Combination.

Let us take a look at some examples to understand how Combinations work:

Problem 1: In how many ways can a committee of 1 man and 3 women can be formed from a group of 3 men and 4 women?

Solution :

No. of ways 1 man can be selected from a group of 3 men =
3 C1 = 3! / 1!*(3-1)! = 3 ways.
No. of ways 3 women can be selected from a group of 4 women = 4C 3 = 4! / (3!*1!) = 4 ways.
Problem 2: Among a set of 5 black balls and 3 red balls, how many selections of 5 balls can be made such that at least 3 of them are black balls.

Solution :

Selecting at least 3 black balls from a set of 5 black balls in a total selection of 5 balls can be:

3 B and 2 R
4 B and 1 R and
5 B and 0 R balls.

Therefore, our solution expression looks like this.
5 C3 * 3C 2 + 5 C4 * 3C 1 + 5 C5 * 3C 0 = 46 ways .
Problem 3: How many 4 digit numbers that are divisible by 10 can be formed from the numbers 3, 5, 7, 8, 9, 0 such that no number repeats?

Solution :

If a number is divisible by 10, its units place should contain a 0.

_ _ _ 0

After 0 is placed in the units place, the tens place can be filled with any of the other 5 digits.

Selecting one digit out of 5 digits can be done in 5C 1 = 5 ways.

After filling the tens place, we are left with 4 digits. Selecting 1 digit out of 4 digits can be done in 4C 1 = 4 ways.

After filling the hundreds place, the thousands place can be filled in 3C 1 = 3 ways.

Therefore, the total combinations possible = 5*4*3 = 60.

A few examples of Combination and Permutation !

Here’s a few examples of combinations (order doesn’t matter) from permutations (order matters).

Combination: Picking a team of 3 people from a group of 10. C(10,3) = 10!/(7! · 3!) = 10 · 9 · 8 / (3 · 2 · 1) = 120.

Permutation: Picking a President, VP and Waterboy from a group of 10. P(10,3) = 10!/7! = 10 · 9 · 8 = 720.

Combination: Choosing 3 desserts from a menu of 10. C(10,3) = 120.

Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. P(10,3) = 720.


** Guys please don’t memorize the formulas, understand why they work. Combinations sound simpler than permutations, and they are. You have fewer combinations than permutations.

Easy Permutation and Combination.

Here’s an easy way to remember what is Permutation and Combination is all about. I hope this can help you out guys :)

Permutation sounds complicated , doesn’t it? And it is. With permutations, every little detail matters. Alice, Bob and Charlie is different from Charlie, Bob and Alice (insert your friends’ names here).

Combinations, on the other hand, are pretty easy going. The details don’t matter. Alice, Bob and Charlie is the same as Charlie, Bob and Alice.

Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter).

A joke: A "combination lock" should really be called a "permutation lock". The order you put the numbers in matters. (A true "combination lock" would accept both 10-17-23 and 23-17-10 as correct.)

Permutations: The hairy details

Let’s start with permutations, or all possible ways of doing something. We’re using the fancy-pants term “permutation”, so we’re going to care about every last detail, including the order of each item. Let’s say we have 8 people:

1: Alice
2: Bob
3: Charlie
4: David
5: Eve
6: Frank
7: George
8: Horatio

How many ways can we award a 1st, 2nd and 3rd place prize among eight contestants? (Gold / Silver / Bronze)

We’re going to use permutations since the order we hand out these medals matters. Here’s how it breaks down:

Gold medal: 8 choices: A B C D E F G H (Clever how I made the names match up with letters, eh?). Let’s say A wins the Gold.

Silver medal: 7 choices: B C D E F G H. Let’s say B wins the silver.

Bronze medal: 6 choices: C D E F G H. Let’s say… C wins the bronze.

We picked certain people to win, but the details don’t matter: we had 8 choices at first, then 7, then 6. The total number of options was 8 · 7 · 6 = 336.

Let’s look at the details. We had to order 3 people out of 8. To do this, we started with all options (8) then took them away one at a time (7, then 6) until we ran out of medals.

We know the factorial is:

Unfortunately, that does too much! We only want 8 · 7 · 6. How can we “stop” the factorial at 5?

This is where permutations get cool: notice how we want to get rid of 5 · 4 · 3 · 2 · 1. What’s another name for this? 5 factorial!

So, if we do 8!/5! we get:

And why did we use the number 5? Because it was left over after we picked 3 medals from 8. So, a better way to write this would be:

where 8!/(8-3)! is just a fancy way of saying “Use the first 3 numbers of 8!”. If we have n items total and want to pick k in a certain order, we get:

And this is the fancy permutation formula: You have n items and want to find the number of ways k items can be ordered:

Combinations, Ho!

Combinations are easy going. Order doesn’t matter. You can mix it up and it looks the same. Let’s say I’m a cheapskate and can’t afford separate Gold, Silver and Bronze medals. In fact, I can only afford empty tin cans.

How many ways can I give 3 tin cans to 8 people?

Well, in this case, the order we pick people doesn’t matter. If I give a can to Alice, Bob and then Charlie, it’s the same as giving to Charlie, Alice and then Bob. Either way, they’re equally disappointed.

This raises an interesting point — we’ve got some redundancies here. Alice Bob Charlie = Charlie Bob Alice. For a moment, let’s just figure out how many ways we can rearrange 3 people.

Well, we have 3 choices for the first person, 2 for the second, and only 1 for the last. So we have 3 · 2 · 1 ways to re-arrange 3 people.

Wait a minute… this is looking a bit like a permutation! You tricked me!

Indeed I did. If you have N people and you want to know how many arrangements there are for all of them, it’s just N factorial or N!

So, if we have 3 tin cans to give away, there are 3! or 6 variations for every choice we pick. If we want to figure out how many combinations we have, we just create all the permutations and divide by all the redundancies . In our case, we get 336 permutations (from above), and we divide by the 6 redundancies for each permutation and get 336/6 = 56.

The general formula is:

which means “Find all the ways to pick k people from n, and divide by the k! variants”. Writing this out, we get our combination formula , or the number of ways to combine k items from a set of n:

Quantitative or Qualitative?

Simple questions for you guys. Try it now :)

Question 1

Determine whether the following statement is about qualitative or quantitative data:

The baby weighs 20 pounds.

Question 2

Determine whether the following statement is about qualitative or quantitative data:

My friend is very happy.

Question 3

Determine whether the following statement is about qualitative or quantitative data:

The sky is greyish-blue.

Question 4

Determine whether the following statement is about qualitative or quantitative data:

Brian is 6 foot 2.

Question 5

Determine whether the following statement is about qualitative or quantitative data:

Fiona has $100.

Discrete or Continuous?

Come on try this out guys :)

Question 1

Determine if the following set of data is discrete or continuous:

The heights of your classmates.

Question 2

Determine if the following set of data is discrete or continuous:

The number of books on your shelves.

Question 3

Determine if the following set of data is discrete or continuous:

The weights of coconuts.

Question 4

Determine if the following set of data is discrete or continuous:

The age of a person.

Question 5

Determine if the following set of data is discrete or continuous:

The number of words in a book.

Statistical Data - Discrete and Continuous

Data can be Descriptive (like "high" or "fast") or Numerical (numbers). So here are some examples to for a deep understanding about Discrete Data and Continuous Data :)

A Numerical Data can be Discrete or Continuous :

Discrete data is counted.
Continuous data is measured.

Discrete Data

Discrete Data can only take certain values.

Example: the number of students in a class (you can't have half a student).

Example: the results of rolling 2 dice:

can only have the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

Continuous Data

Continuous Data can take any value (within a range)
Examples:

• A person's height: could be any value (within the range of human heights), not just certain fixed heights,

• Time in a race: you could even measure it to fractions of a second,

• A cat's weight,

• The length of a leaf,

Lots more!

Statistical Data - Quantitative and Qualitative

There are two general types of data. 

Quantitative data is information about quantities that is, information that can be measured and written down with numbers. Some examples of quantitative data are your height, your shoe size, and the length of your fingernails. Speaking of which, it might be time to call Guinness. You've got to be close to breaking the record.

Qualitative data is information about qualities information that can't actually be measured. Some examples of qualitative data are the softness of your skin, the grace with which you run, and the color of your eyes. However, try telling Photoshop you can't measure color with numbers.

Here's a quick look at the difference between qualitative and quantitative data.

The age of your car. (Quantitative.)

The number of hairs on your knuckle. (Quantitative.)

The softness of a cat. (Qualitative.)

The color of the sky. (Qualitative.)

The number of pennies in your pocket. (Quantitative.)


** Remember, if we're measuring a quantity, we're making a statement about quantitative data. If we're describing qualities, we're making a statement about qualitative data. Keep your L's and N's together and it shouldn't be too tough to keep straight.

Find Mean, Mode and Median for Grouped data

The Histogram shows an even spread of data, indicating that sometimes the Coffee Shop is very busy, while other times they are making less than eight cappuccinos per hour.We now want to find the Average Number of Cappuccinos made every hour.

There are three types of Averages:

The Mean
The Median
The Mode

In this topic we calculate all three of these averages for the coffee shop example.

Finding the Range

The “Range” is the easiest Statistic to determine for Grouped Data. We simply take the end of the Highest Interval, and subtract the Beginning of the first Interval.

Range = Maximium – Minimum

For our Coffee Statistics, the Highest Group is 16-19, so our High Value “Maximum” is 19. The Lowest Group is 0-3, so the Low Value “Minimum” is zero.

Range = Maximium – Minimum = 19 – 0 = 19
The Range can also be stated as “0 to 19″



The Modal Class

The “Mode” is what happens most of the time, or on most occassions.

The “Mode” is the simplest Grouped Average to find.
It can be read straight from the Frequency Table, or straight from the Graph.

Sometimes we have more than one Group which is the most popular.

In these situations, we can have a two modes or a “Bimodal” situation, or three modes which is called “Trimodal”.

The Median Class

Finding the Median Class involves some working out steps to be applied to our original Frequency Table. There are three Main Steps:

1) Finding the half-way midpoint in the Frequency values.

2) Adding a third column to our Frequency Table where we calculate “Cumulative Frequency” values.

3) Locating the half-way point in the Cumulative Frequency Column, and then seeing which Class Interval lines up with this half-way point.

How we do each of these steps is as follows.

There are two ways to find the half-way midpoint in the Frequency values.

We can either write out the numbers from 1 to the Total frequency value and manually find the middle; or we can use a simple math formula to find this value.

Rather than writing out a long list of numbers, It is much easier to use the formula: 

Middle = Total Frequency + 1 and then divide by 2.

Find Mean, Mode, Median with Frequency Table

Find the mean, median and mode for this grouped data of test scores.

Scores   Frequency
   65                2
   70                3
   75                2
   80                5
   85                8
   90                7
   95                5
   100              3

This problem could be solved by entering ALL 35 scores into one list, with the score 65 appearing twice, 70 appearing 3 times, and so on.  But if we deal with the data as we deal with a frequency histogram, we can accomplish our task more quickly.

Enter the scores into L 1 and the frequencies into L 2 .
(See Basic Commands for entering data.)

Find the Mean and Median:

Method 1: (fast and easy )
Press 2nd MODE (QUIT) to return to the home screen.
Press 2nd STAT (LIST). Arrow to the right to
MATH.
Choose option #3: mean( if you want the mean.

Choose option #4: median( if you want the median.

Your choice will appear on the home screen waiting for you to tell it which list to use. This time we will tell the calculator the list containing the scores AND the list containing the frequencies. Notice the comma separating the lists.

Remember the List names appear on the face of the calculator above the number keys 1-6.

Find the Mean and Median:

Method 2: (a bit more sophisticated)
Press STAT. Arrow to the right to CALC. Now choose option #1: 1-Var Stats .

When 1-Var Stats appears on the home screen, tell the calculator the name of the list containing the scores AND the name of the list containing the frequency
(such as: 1-Var Stats L1, L2)

Press ENTER.

Arrow up and down the screen to see the statistical information about the data.

Find the Mode: The mode can be quickly determined by examining the chart to see which score occurred most often. No calculator work needed.
The mode is 85.

Test: Mean, Mode and Median

Questions and Answers

1. Find the mean in the following numbers:19, 21, 18, 17, 18, 22, 46

A. 23
B. 16
C. 19
D. 500
None of the above

2. What is the median in the following numbers:19, 21, 18, 17, 18, 22, 46

A. 15
B. 19
C. 21
D. 22
None of the above

3. What does mean, mean?

4. What does median mean?

5. What does mode mean?

6. What does range mean?

7. Find the mean in the following numbers:9, 8, 15, 8, 20

A. 12
B. 16
C. 15
D. 10
None of the above

8. Find the median in the following numbers:9, 8, 15, 8, 20

A. 20
B. 15
C. 9
D. 22
None of the above

9. Find the mode in the following numbers:9, 8, 15, 8, 20

A. 15
B. 8
C. 9
D. 20
None of the above

10. Find the range in the following numbers:9, 8, 15, 8, 20

A. 12
B. 900
C. 14
D. 15
None of the above

11. Find the mean in the following numbers:36, 38, 33, 34, 32, 30, 34, 35

A. 54
B. 34
C. 74
D. 44
None of the above

12. Find the median in the following numbers:36, 38, 33, 34, 32, 30, 34, 35

A. 34
B. 55
C. 23
D. 45
None of the above

13. Find the mode in the following numbers:36, 38, 33, 34, 32, 30, 34, 35

A. 34
B. 38
C. 36
D. 30
None of the above

More examples for you :)

Okay guys, here i posted more examples about the
measures of central tendency :)

Example No.1

Find the mean, median and mode for the following data:

5, 15, 10, 15, 5, 10, 10, 20, 25, 15

(You will need to organize the data.)

5, 5, 10, 10, 10, 15, 15, 15, 20, 25

Mean: Sum of all data ÷ No.of terms


130 / 10  = 13 (ans)

Median: 5, 5, 10, 10, 10, 15, 15, 15, 20, 25

**  Listing the data in order is the easiest way to find the median.

The numbers 10 and 15 both fall in the middle.
Average these two numbers to get the median.
10 + 15 =12.5 / 2
               = 6.25 (ans)

Mode: Two numbers appear most often: 10 and 15.

There are three 10's and three 15's.

In this example there are two answers for the mode.


Example No.2

For what value of x will 8 and x have the same mean (average) as 27 and 5?

First, find the mean of 27 and 5:
27 + 5 = 16 / 2

Now, find the x value, knowing that the average of x and 8 must be 16:

x + 8 ÷ 2 = 16

32 = x + 8 cross multiply

-8 - 8

= 24 (ans)


Example No.3

On his first 5 biology tests, Bob received the following scores: 72, 86, 92, 63, and 77. What test score must Bob earn on his sixth test so that his average (mean score) for all six tests will be 80? Show how you arrived at your answer.

Possible solution:

Set up an equation to represent the situation. Remember to use all 6 test scores:
72 + 86 + 92 + 63 + 77 + x = 80 / 6

cross multiply and solve: (80)(6) = 390 + x

480 = 390 + x

- 390   -390

90 = x

Bob must get a 90 on the sixth test.


Example No.4

The mean (average) weight of three dogs is 38 pounds. One of the dogs, Sparky, weighs 46 pounds. The other two dogs, Eddie and Sandy, have the same weight. Find Eddie's weight.

Let x = Eddie's weight ( they weigh the same, so they are both represented by " x ".)
Let x = Sandy's weight

Average:   sum of the data divided by the number of data.

x + x + 46 = 38 cross multiply and solve
3(dogs)
(38)(3) = 2x + 46
114 = 2 x + 46

-46 -46
68 = 2x
2 2
34 = x Eddie weighs 34 pounds.

Do it yourself !

Here I posted a simple question on finding the Mean, Mode, Median and Range. Yet, it might be easy but once you do it careless, you might get the answer wrong :)

Stated below is the age of 8 siblings.

Hanna 67

Marissa 50

Mark 50

Brian 44

Mick 36

Kim 34

Rose 32

Nora 30

• Noted, Marissa and Mark are twins :)

Find,

(a) Mean

(b) Mode

(c) Median

(d) Range

WARM UP question !

Come on guys, try this out ! 

The list gives the scores a group of students who took part in a mathematics quiz.

10, 7, 14, 38, 26, 16, 10, 21

Find:


(a) Mode

= 10 (the most popular number)

(b) Mean

:  10 + 7 + 14 + 38 + 26 + 16 + 10 + 21 / 8 = 17.75 (Plus all the numbers and then divided by how many terms)

(c) Median

: Arranged the numbers from the smallest to biggest

7 , 10 , 10, 14 , 16, 21, 26, 38

Since there is 14 and 16 are the middle numbers. Plus 14 + 16 then divided it into 2. The answer is 15.

(d) Range

: This will be the simplest one. Just subtract the highest number with the smallest one.

38 - 7 = 31 (ans)

(e) The lower and upper quartile

: there is three quartile altogether

• lower quartile
• middle quartile
• upper quartile

So to find the:

Lower quartile -> 1/4 × 8 (no.of terms)
= 2.

Upper quartile -> 3/4 × 8 (no.of.terms)
= 6.

Lower = 10.
Upper = 26.

(d) The inter quartile range

: Minus the upper quartile - lower qurtile

26 - 10 = 16 (ans)

Deep understanding about Measures of central tendency :)

  

Understand more about mean, mode and median :)

Definitions:

Mean - When people say "average" they usually are talking about the mean. You can figure out the mean by adding up all the numbers in the data and then dividing by the number of numbers. For example, if you have 12 numbers, you add them up and divide by 12. This would give you the mean of the data. 

Mode - The mode is the number that appears the most. There are a few tricks to remember about mode: 

If there are two numbers that appear most often (and the same number of times) then the data has two modes. This is called bi-modal. If there are more than 2 then the data would be called multi-modal. If all the numbers appear the same number of times, then the data set has no modes. 

Median - The median is the middle number of the data set. It is exactly like it sounds. To figure out the median you put all the numbers in order (highest to lowest or lowest to highest) and then pick the middle number. If there is an odd number of data points, then you will have just one middle number. If there is an even number of data points, then you need to pick the two middle numbers, add them together, and divide by two. That number will be your median. 

Practice MEAN, MEDIAN & MODE :)

Here are two practice questions for you guys. Easy and simple one to understand :)

1. The weekly salaries of six employees at Jolibee are $160, $280, $100, $170, $150, $250.  For these six salaries, find:

(a) the mean 

(b) the median 

(c) the mode

List the data in order: 100, 150, 150, 180, 250, 280

                                                                     
Mean:    100+ 150+ 150+ 180 + 250 + 280 =185 (ans)
                                        6                          

Median:    90,150,150,180,250,280
The two numbers that fall in the middle need to be
averaged.     150 + 180 (Answer divided by 2) = 165 (ans)

Mode:   The number that appears the most is = 150 (ans)

2. Adrian has grades of 84, 65, and 76 on three math tests.  What grade must he obtain on the next test to have an average of exactly 80 for the four tests?

84 + 65 + 76+ x   =   80 (average) cross multiply 
                   4                                         and solve

                        (4)(80) = 225 + x
                             320 = 225 + x
                           -225   -225

                               95 = x 

               Adrian needs a 95 on his next test.

** Try not to memorize the formulas. It is better for you to understand the step by step process in order to find the correct answer.