What is the number of possible words that can be made using the word easy quiz such that vowels always come together?

Ever heard of typoglycemia? Even if you haven’t, chances are you’ll recognize one of the viral puzzles used to demonstrate the phenomenon. Starting around 2003, an email began to circulate claiming that scrambled English words are just as easy to read as the original words.

However, as interesting as the original email was, it didn’t actually tell the whole truth. There’s more to scrambled words than meets the eye.

What is typoglycemia?

That viral email tested our ability to read scrambled words. Here’s what it looks like:

Aoccdrnig to a rscheearch at Cmabrigde Uinervtisy, it deosn’t mttaer in waht oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist and lsat ltteer be at the rghit pclae. The rset can be a toatl mses and you can sitll raed it wouthit porbelm. Tihs is bcuseae the huamn mnid deos not raed ervey lteter by istlef, but the wrod as a wlohe.

Could you read it? Even with a mistake in this viral email (the letters in rscheearch cannot spell researcher), the truth is that most fluent English speakers can read and understand it.

The word-scrambling phenomenon has a punny name: typoglycemia, playing with typo and glycemia(the condition of having low blood sugar). Typoglycemia can refer to to the phenomenon in which words can be read despite being jumbles, or it can refer to the ability to read such texts. Still, though the word may sometimes be referenced in actual research, it’s not a formal term, nor is it all that commonly used outside the context of such memes.

Is typoglycemia real or a trick?

Does it take you nanoseconds to solve a Word Jumble? No? While your brain can breeze through some word scrambles, it’s more complicated than that viral email suggests.

Matt Davis, a researcher at the MRC Cognition and Brain Sciences Unit at Cambridge University, helped us sort it out. Here’s what Davis believes the email got right: unless you have a rare brain disorder, people read words as whole units, not letter-by-letter. That’s one of the factors explaining why we can “magically” read the message.

There’s a reason we call them “sight words.” Here are some great sight word activities to help your child learn to read and write.

But here’s where Davis reminds us why the daily Word Jumble still manages to scramble our brains for breakfast. That trending email led us to believe all we need is for “the first and last letters to be in the right place” and nothing else matters. However, it’s much more complicated than that.

What makes a scrambled word easier to read?

Here are some other factors a jumbled passage needs in order for most people to easily read it:

1. The words need to be relatively short.
2. Function words (be, the, a, and other words that provide grammatical structure) can’t be jumbled, or else the reader will likely struggle.
3. Switching (or transposing) the letters makes a big difference. Letters beside each other in a word can be switched without much creating much difficulty for the reader. When letters farther apart are switched, it’s harder. For example, look at porbelm vs. pelborm (for “problem”).
4. We understand scrambled words better when their sounds are preserved: toatl vs. talot (for “total”).
5. Here’s a big one: the passage is readable because it’s predictable—that is, the topic is logically explained, with context providing very good clues about what words will be used.

Other factors play into it as well, like preserving double letters. For example, in the word according, the scrambled email keeps the cc intact (“aoccdrnig”). Double letters are contextual markers that give good hints. When we scramble it up in another way (such as “ancdircog”), it can be much less recognizable.

All told, we’re code-making machines (we speak the code of English) and we’re wired to find meaning in text, in part by looking at contextual cues. However, the codes can only be scrambled so much before we get lost.

Research shows that typos definitely interfere with reading speed. (There’s a reason we have spell-checkers!) Tricky jumble puzzles that can take hours to complete also prove that, in the end, letter order and spelling can absolutely make or break our comprehension of a word.

In this article you’ll learn about Permutation and Combination problems: Definition, formulas, solved examples and a quiz with practice questions.

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, nPn = 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:
nPr = 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 = 4P3 = 24 ways.

After 3 vowels take 3 places, no. of ways 4 consonants can take 4 places = 4P4 = 4! = 24 ways.

Therefore, total number of permutations possible = 24*24 = 576 ways.

Combinations

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 nCr is

nCr = 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

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

Important Combination formulas

nCn = 1

nC0 = 1

nC1 = n

nCr = nC(n-r)

The number of selections possible with A, B, C, taken all at a time is 3C3 = 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 = 3C1 = 3! / 1!*(3-1)! = 3 ways.

No. of ways 3 women can be selected from a group of 4 women = 4C3 = 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.
5C3 * 3C2 + 5C4 * 3C1 + 5C5 * 3C0 = 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 5C1 = 5 ways.

After filling the tens place, we are left with 4 digits. Selecting 1 digit out of 4 digits can be done in 4C1 = 4 ways.

After filling the hundreds place, the thousands place can be filled in 3C1 = 3 ways.

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

Permutations and Combinations Quiz

Try these practice problems.

Problem 1: Click here

Answer 1: Click here

Problem 2: Click here

Answer 2: Click here

Problem 3: Click here

Answer 3: Click here

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