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!