In how many ways can letters in the word "computer" be arranged?
such that all the vowels are together.
such that all the vowels are together.
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please can you provide the logic behind this. Please show how you arrived at this.kanwar86 wrote:6!3! ways
hi lots of thankskanwar86 wrote:Number of vowels are 3 and they are all different...each vowel being represented by |
Number of consonants are 5 in number (all different)..each consonant represented by 0 (though they are all different..bear wid me)
Now, we want all vowels to be together
that is., |||00000 (this kind of structure)
So, what we do is we tie up these three vowels together and make it a single object (in square bracket) just like any other consonant [|||]00000, 0[|||]0000, 00[|||]000, 000[|||]00 etc.
So, in effect we have (8(total alphabets)-3(vowels to be kept together)+1(vowels are combined together to form a single object))= 6 different consonants which can be arranged in 6! ways.
But, that object in square brackets is composed of 3 vowels which can be mutually arranged in 3! ways.
Hence, the total number of ways = 6!*3!
Hope that helped.
The difference lies in the order of appearance of vowels in the word "Double"gmatter2012 wrote:hi lots of thankskanwar86 wrote:Number of vowels are 3 and they are all different...each vowel being represented by |
Number of consonants are 5 in number (all different)..each consonant represented by 0 (though they are all different..bear wid me)
Now, we want all vowels to be together
that is., |||00000 (this kind of structure)
So, what we do is we tie up these three vowels together and make it a single object (in square bracket) just like any other consonant [|||]00000, 0[|||]0000, 00[|||]000, 000[|||]00 etc.
So, in effect we have (8(total alphabets)-3(vowels to be kept together)+1(vowels are combined together to form a single object))= 6 different consonants which can be arranged in 6! ways.
But, that object in square brackets is composed of 3 vowels which can be mutually arranged in 3! ways.
Hence, the total number of ways = 6!*3!
Hope that helped.
here is a similar question
In how many ways can the letters of the word "double" be rearranged such that the order in which the vowels appear does not change?
This question says that they must appear in the same order and not together.how will the solution change , please do explain the logic.
I posted a solution here:gmatter2012 wrote: In how many ways can the letters of the word "double" be rearranged such that the order in which the vowels appear does not change?
This question says that they must appear in the same order and not together.how will the solution change , please do explain the logic.
hi many many thankskanwar86 wrote:The difference lies in the order of appearance of vowels in the word "Double"gmatter2012 wrote:hi lots of thankskanwar86 wrote:Number of vowels are 3 and they are all different...each vowel being represented by |
Number of consonants are 5 in number (all different)..each consonant represented by 0 (though they are all different..bear wid me)
Now, we want all vowels to be together
that is., |||00000 (this kind of structure)
So, what we do is we tie up these three vowels together and make it a single object (in square bracket) just like any other consonant [|||]00000, 0[|||]0000, 00[|||]000, 000[|||]00 etc.
So, in effect we have (8(total alphabets)-3(vowels to be kept together)+1(vowels are combined together to form a single object))= 6 different consonants which can be arranged in 6! ways.
But, that object in square brackets is composed of 3 vowels which can be mutually arranged in 3! ways.
Hence, the total number of ways = 6!*3!
Hope that helped.
here is a similar question
In how many ways can the letters of the word "double" be rearranged such that the order in which the vowels appear does not change?
This question says that they must appear in the same order and not together.how will the solution change , please do explain the logic.
That is, o should come before u, and u should come before e.
Hence, if we permute these three in a group only one combination follows the desired order.
Now we have 6 alphabets which can be re arranged in 6! ways
Out of 6! ways, there are 3! ways in which those three vowels have been rearranged.
But we are looking for permutations where that "one" order (discussed above) is being followed.
So, we divide 6!/3! = 720/6 = 120 ways (Ans)
Mitch thank youGMATGuruNY wrote:I posted a solution here:gmatter2012 wrote: In how many ways can the letters of the word "double" be rearranged such that the order in which the vowels appear does not change?
This question says that they must appear in the same order and not together.how will the solution change , please do explain the logic.
https://www.beatthegmat.com/tough-permut ... 20042.html
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