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Re: In how many ways can the letters of the word MANIFOLD be arr
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06 Feb 2014, 00:38

another way to solve would be to find out the total ways to arrange the 8 letters and deduct those choices where the vowels are always together.

so total number of ways to arrange 8 letters would be 8!

choices where vovels are always together would be when you treat 3 vowels as one letter and arrange the remaining 5 letters to give 6! ways to write the letters where the vowels are always together.

the vowels can be further arranged in 3! ways.

so we have 8! - 6!*3! which gives 36000

please prove this wrong so that the flaw in the logic is detected.

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06 Feb 2014, 00:47

1

ratnanideepak wrote:

In how many ways can the letters of the word MANIFOLD be arranged so that the vowels are separated ?

A. 14400 B. 36000 C. 18000 D. 24000 E. 22200

another way to solve would be to find out the total ways to arrange the 8 letters and deduct those choices where the vowels are always together.

so total number of ways to arrange 8 letters would be 8!

choices where vovels are always together would be when you treat 3 vowels as one letter and arrange the remaining 5 letters to give 6! ways to write the letters where the vowels are always together.

the vowels can be further arranged in 3! ways.

so we have 8! - 6!*3! which gives 36000

please prove this wrong so that the flaw in the logic is detected.

The point is that {total} - {all three vowels together} does not give the cases where all the vowels are separated: you still get the cases where any two of them are together. For example, {AI}MNFOLD or MAN{IO}FLD ...

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06 Feb 2014, 01:31

so how do you remove any two of them together from this total ?....i just want to arrive at the correct answer by viewing the entire logic by this method....do we have to deduct 7! * 2 further by treating any two vowels as one...if so the answer is till not the same....

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06 Feb 2014, 01:42

3

1

ratnanideepak wrote:

In how many ways can the letters of the word MANIFOLD be arranged so that the vowels are separated ?

A. 14400 B. 36000 C. 18000 D. 24000 E. 22200

so how do you remove any two of them together from this total ?....i just want to arrive at the correct answer by viewing the entire logic by this method....do we have to deduct 7! * 2 further by treating any two vowels as one...if so the answer is till not the same....

*M*N*F*L*D*

Exactly two of the vowels are together. Consider the two vowels as one unit: {X, Y}

The # of ways to choose which two vowels out of three are together = \(C^2_3=3\) The # of ways to arrange these two within their unit = 2!; The # of ways to choose an empty slot for that unit = 6; The # of ways to choose an empty slot for the third vowel = 5. The # of ways to arrange MNFLD = 5!.

Re: In how many ways can the letters of the word MANIFOLD be arr
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06 Feb 2014, 03:47

Bunuel wrote:

ratnanideepak wrote:

In how many ways can the letters of the word MANIFOLD be arranged so that the vowels are separated ?

A. 14400 B. 36000 C. 18000 D. 24000 E. 22200

so how do you remove any two of them together from this total ?....i just want to arrive at the correct answer by viewing the entire logic by this method....do we have to deduct 7! * 2 further by treating any two vowels as one...if so the answer is till not the same....

*M*N*F*L*D*

Exactly two of the vowels are together. Consider the two vowels as one unit: {X, Y}

The # of ways to choose which two vowels out of three are together = \(C^2_3=3\) The # of ways to arrange these two within their unit = 2!; The # of ways to choose an empty slot for that unit = 6; The # of ways to choose an empty slot for the third vowel = 5. The # of ways to arrange MNFLD = 5!.

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07 Feb 2014, 03:42

WoundedTiger wrote:

Bunuel wrote:

ratnanideepak wrote:

In how many ways can the letters of the word MANIFOLD be arranged so that the vowels are separated ?

A. 14400 B. 36000 C. 18000 D. 24000 E. 22200

so how do you remove any two of them together from this total ?....i just want to arrive at the correct answer by viewing the entire logic by this method....do we have to deduct 7! * 2 further by treating any two vowels as one...if so the answer is till not the same....

*M*N*F*L*D*

Exactly two of the vowels are together. Consider the two vowels as one unit: {X, Y}

The # of ways to choose which two vowels out of three are together = \(C^2_3=3\) The # of ways to arrange these two within their unit = 2!; The # of ways to choose an empty slot for that unit = 6; The # of ways to choose an empty slot for the third vowel = 5. The # of ways to arrange MNFLD = 5!.

2. The number of arrangements of six units {AIO }{M}{N}{F}{L}{D} is indeed 6!*3! but you don't need further to multiply this by 6, because 6!*3! already gives all the possible arrangements of {AIO }{M}{N}{F}{L}{D}.

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21 Feb 2016, 10:19

3

i got to the answer choice slightly differently...and I dont know if it's a correct way to solve it... we have *C*C*C*C*C* 5 consonants, and 6 places for vowels. consonants can be arranged in 5! ways, or 120 ways. 1st vowel can be arranged in 6C1 ways 2nd vowel can be arranged in 5C1 ways 3rd vowel can be arranged in 4C1 ways

now multiply everything: 5!*6C1*5C1*4C1 = 120*6*5*4 = 14,400

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01 Sep 2017, 07:19

Top Contributor

ratnanideepak wrote:

In how many ways can the letters of the word MANIFOLD be arranged so that the vowels are separated ?

A. 14400 B. 36000 C. 18000 D. 24000 E. 22200

Take the task of arranging the 8 letters and break it into stages.

Stage 1: Arrange the 5 CONSONANTS (M, N, F, L and D) in a row We can arrange n unique objects in n! ways. So, we can arrange the 5 consonants in 5! ways (= 120 ways) So, we can complete stage 1 in 120 ways

IMPORTANT: For each arrangement of 5 consonants, there are 6 spaces where the VOWELS can be placed. For example, in the arrangement MNDLF, we can add spaces as follows _M_N_D_L_F_ So, if we place each vowel in one of the available spaces, we can ENSURE that the vowels are separated.

Stage 2: Select a space to place the A. There are 6 spaces to choose from, so we can complete stage 2 in 6 ways.

Stage 3: Select a space to place the I. There are 5 remaining spaces to choose from, so we can complete stage 3 in 5 ways.

Stage 4: Select a space to place the O. There are 4 remaining spaces to choose from, so we can complete stage 4 in 4 ways.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus arrange all 8 letters) in (120)(6)(5)(4) ways (= 14,400 ways)

Answer: A

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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