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# for the magic word how many different types of arrangement are possible so that the vowels are always together?

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### Mohammed

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## Permutation and Combination

This is the aptitude questions and answers with discussion section on "Permutation and Combination" with explanation for various interview, competitive examination and entrance test. Solved examples with detailed answer description, explanation are given and it would be easy to understand - Discussion page for Q.677.

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## Aptitude :: Permutation and Combination - Discussion

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### Discussion :: Permutation and Combination - General Questions (Q.No.2)

Permutation and Combination - Important Formulas

«« Permutation and Combination - General Questions

2.

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

[A]. 360 [B]. 480 [C]. 720 [D]. 5040 [E]. None of these Answer: Option C Explanation:

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Video Explanation: https://youtu.be/WCEF3iW3H2c

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Shoba said: (Sep 4, 2010)

Other than vowels there are only 4 letter then how it s possible to get 5!.

Sai said: (Sep 10, 2010)

HI SHOBA,.

4 consonants + set of vowels (i. E. , L+N+D+G+ (EAI) ).

We should arrange all these 5. So we get 5!.

I think you understood.

Sachin Kumar said: (Sep 29, 2010)

Please make me understand this answer as did not get.

Sri said: (Oct 10, 2010)

As vowels are together take (EAI) as single letter i.e. , total no of letters are 5 (L, N, D, G, {EAI}).

No of ways can arrange these 5 letters are 5! ways.

Now we arranged 5 letters (L, N, D, G, {EAI}).

Next we have to arrange E, A, I (they may be EAI/EIA/AEI/AIE/IAE/IEA).

All these combinations imply that vowels are together.

So we have to multiply 5! and 3!.

Subbu said: (Jan 6, 2011)

7!=5040

Madhusudan said: (Mar 14, 2011)

Could you kindly let me know what is ( ! ).Howe 5! = 120 ?

Sundar said: (Mar 14, 2011)

5! = 5 Factorial = 1 x 2 x 3 x 4 x 5 = 120

Laxmikanth said: (Mar 26, 2011)

When should we take the one or more letters as a single unit and why?

Moaned said: (Apr 8, 2011)

I am not getting

Jessie said: (Apr 10, 2011)

7 letter word = LEADING

CONDITION = VOWELS TO BE TOGETHER, HENCE (EAI) TO FOR A WORD

SO NO. OF WORDS = L,(EAI),D,N,G = 5

permuation to arrange 5 letters = nPr= n!/(n-r)!=5!/0!=5!

0! is assumed to be 1!)

EAI can be arranged among each other in = nPr = 3!/(3-3)= 3!

hence 5! x 3! = 120 x 6 = 720

Mayur said: (Apr 15, 2011)

How it came like nPr formula and how you have solve it? please let me know.

Bharti said: (Jul 11, 2011)

Thanks a lot. I had confusion before your explanations. Thanks a lot.

[email protected] said: (Aug 5, 2011)

@mayur.

You just look when ever you open the new exercise there will be availability of basic formulas. If you go through them half of the task would be finished easily. Have a good day buddy.

Siva said: (Oct 16, 2011)

Permutations and combinations always make me to confuse much.

How to decide based on the descriptive aptitude question ?

Shiva said: (Oct 18, 2011)

Why we have taken 5 (4 + 1) ?

Santosh Kumar Pradhan said: (Oct 27, 2011)

Total member 7

out of which 5 consonant and 3 vowels

take 3 vowls as a 1

hen number of consonant will arrange 5! ways

and these 3 vowels will arrange 3! ways

though,5!*3!=720

Bhavik said: (Nov 4, 2011)

How to read these nPr ?

Sagar Choudhary said: (Nov 6, 2011)

What is this 5! and 3! ?

A.Vamsi Krishna said: (Jan 30, 2012)

"!" this implies factorial that means a number is multiplied

like for example take number 5 then its factorial will be

taken as 5*4*3*2*1 and this is equal to 120.

Ncs said: (Feb 2, 2012)

Why not 5! + 3! ?

Raj said: (May 21, 2012)

Friends, How do you say this question is permutation.

Mahanthesh said: (Jul 30, 2012)

Hi guys. As per the question, it s mentioned that all vowels should be together, but it has not been mentioned that it should be EAI. According to me these 3 vowels can be arranged in 3*2=6 ways. And the remaining letters LDNG can be arranged in 4*3*2*1= 24 ways. Can anyone explain me about this please ?

Sandeep said: (Sep 7, 2012)

can any one explain ? y 5!*3! & y not like this 5!+3!

Srk said: (Oct 9, 2012)

@Mahantesh.

1. There is given a word LEADING in this LDNG (consonents) , EAI (vowels).

2. They asked here vowels always come together and so we should have to take LDNG (EAI).

3. We can take as (4+1) ! i.e. here we have to take vowels as together as 1. So we have a chance of 5!.

4. But with in vowels we have many arrangements i.e 3!

5. Finally 5!*3!

Shafi said: (Oct 11, 2012)

@Mahantesh,

You are correct 3! and 4!, because you spitted vowels and consonants,

But to combine them vowels+consonants.

i.e. (4 consonants + 1 set of vowels) = 5!

And (3 vowels {1 set}) = 3!

So 5!*3! = 5x4x3x2x1x3x2x1=720.

Adhityasena said: (Feb 9, 2013)

I get the answer to be 2*720. Because, since the vowels must come together in the word LEADING, which actually has 7 letters, E and A can be taken as one unit, so now we have L, EA, D, I, N, G to be arranged and which can be done in 6! ways. But among E and A there are 2 arrangements namely EA and AE, So the final answer is 2*720. Am I right !

स्रोत : www.indiabix.com

## In how many different ways can the letters of the word TRAINER be arranged so that the vowels always come together?A. 1440B. 120C. 720D. 360

In how many different ways can the letters of the word TRAINER be arranged so that the vowels always come together?A. 1440B. 120C. 720D. 360. Ans: Hint: To solve this problem we have to know about the concept of permutations and combinations. But her...

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

A. 1440 B. 120 C. 720 D. 360 Answer Verified 215.7k+ views 1 likes

Hint: To solve this problem we have to know about the concept of permutations and combinations. But here a simple concept is used. In any given word, the number of ways we can arrange the word by jumbling the letters is the number of letters present in the word factorial. Here factorial of any number is the product of that number and all the numbers less than that number till 1.

⇒n!=n(n−1)(n−2).......1

⇒n!=n(n−1)(n−2).......1

Complete step by step answer:

Given the word TRAINER, we have to arrange the letters of the word in such a way that all the vowels in the word TRAINER should be together.

The number of vowels in the word TRAINER are = 3 vowels.

The three vowels in the word TRAINER are A, I, and E.

Now these three vowels should always be together and these vowels can be in any order, but they should be together.

Here the three vowels AIE can be arranged in 3 factorial ways, as there are 3 vowels, as given below:

The number of ways the 3 vowels AIE can be arranged is =

3! 3!

Now arranging the consonants other than the vowels is given by:

As the left out letters in the word TRAINER are TRNR.

The total no. of consonants left out are = 4 consonants.

Now these 4 consonants can be arranged in the following way:

As in the 4 letters TRNR, the letter R is repeated for 2 times, hence the letters TRNR can be arranged in :

⇒ 4! 2! ⇒4!2!

But the letters TRNR are arranged along with the vowels A,I,E, which should be together always but in any order.

Hence we consider the three vowels as a single letter, now TRNR along with AIE can be arranged in:

⇒ 5! 2! ⇒5!2!

But here the vowels can be arranged in

3! 3!

as already discussed before.

Thus the word TRAINER can be arranged so that the vowels always come together are given below:

⇒ 5! 2! ×3!= 120×6 2 ⇒5!2!×3!=120×62 ⇒360 ⇒360

The number of ways the word TRAINER can be arranged so that the vowels always come together are 360. Note: Here while solving such kind of problems if there is any word of

n n

letters and a letter is repeating for

r r

times in it, then it can be arranged in

n! r! n!r!

number of ways. If there are many letters repeating for a distinct number of times, such as a word of

n n letters and r 1 r1 repeated items, r 2 r2 repeated items,……. r k rk

repeated items, then it is arranged in

n! r 1 ! r 2 !...... r k ! n!r1!r2!......rk! number of ways.

स्रोत : www.vedantu.com

## How many different types of arrangements are possible for the word MAGIC so that the vowels are always together?

Answer (1 of 4): Start by considering the pair of vowels as a single unit V. Now you’re arranging the four units “M”, “G”, “C” and “V”. There are 4!=24 ways to do that. Now all you need to do is to notice there are two different versions of each of these permutations: one in which the vowel grou...

How many different types of arrangements are possible for the word MAGIC so that the vowels are always together?

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Author has 793 answers and 134.8K answer views1y

MAGIC = MGC+AI

Consider AI as 1 unit. We now have 4 letters that can be arranged in 4! ways. AI can be arranged amongst themselves in 2 ways (AI, IA).

Req. arrangements = 2*4!

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Start by considering the pair of vowels as a single unit V. Now you’re arranging the four units “M”, “G”, “C” and “V”. There are 4!=24 ways to do that.

Now all you need to do is to notice there are two different versions of each of these permutations: one in which the vowel group V is “AI”, the other where it is “IA”. This doubles the number of options, giving 48 in all.

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Mario Skrtic

IT at Public Libraries (2003–present)Author has 1.5K answers and 336.2K answer views1y

AI or IA are together and act as one.

We have M, G, C and AI or IA.

The answer is 2*4!=48

Ellis Cave

40+ years as an Electrical EngineerAuthor has 5.2K answers and 2.8M answer views1y

Using the J programming language, brute force approach:

(J (programming language) - Wikipedia

)

+/m=.1 1 ssmb"1 n=.(p=.perm 5){0 1 0 1 0

48

The answer is 48 different arrangements.

List them:

Will Scathlocke Upvoted by Steve Rapaport

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Related

Are there any words without any vowels?

Depends on how you define a “vowel”. If you define a vowel graphically (i.e. as something written) as they do on Wheel of Fortune (and that’s a perfectly valid way of doing it), i.e. as “a”, “e”, “i”, “o”, and “u”, then even in English there are words without vowels. “Rhythm” is one; “hymn” is another.

If you define a vowel phonically, but still exclude semi-vowels (/w/ and /y/) and sonorants such as (/r/, /l/, /m/, /n/), then various languages have words without vowels. For example, there is this Czech tongue-twister:

strc prst strz krk (the “c” should have a hook over it)

“stick a finger throug

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In how many ways can the letters of the word ‘rainbow’ be arranged so that only two vowels always come together?

2880

There are three vowels a, i, and o. Two vowels can be selected 3 ways. Let us first take a and i. Consider these two vowels as one letter. Now there are 6 letters. They can be arranged in 6!*2! = 1440. As there are three cases, the total arrangements will be 4320.

Now these arrangements will also include all the three vowels together.

‘O’ can come before or after a and i. ‘oai’, ‘aio’, ‘oia’, and ‘iao’. All these fours cases are distinct. Similarly, we have another 8 distinct cases.

Such number of arrangements is 2*5!*3! =1440.

Hence, the required answer is 4320 - 1440 = 2880.

Alternatively, co Tim Farage

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In how many ways can the letters of the word “element” be arranged so that the vowels are always together?

In math, whenever you are counting the number of arrangements of symbols, this is called a permutation problem.

So you question could have also been asked like this: “How many permutations are there of the letters of the word ‘element’, such that the vowels are always together?”

If these 7 letters were all different, the answer would be 7!.

But since the three e’s must be together, think of them as a single symbol ‘eee’.

So now we want to know how many ways we can permute the symbols ‘eee’, ‘l’, ‘m’, ’n’ and ‘t’.

There are 5 symbols here, so we can permute them in 5! ways.

Since 5! = 120, it would b

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