In how many ways can the letters of the word MANIFOLD be arr

We have 3 vowels: AIO and 5 consonants: MNFLD.

Consider the following case: *M*N*F*L*D* If we place vowels in any 3 empty slots (*) then all vowels will be separated by at least one consonant:

The # of ways to choose 3 empty slots out of 6 for 3 vowels = \(C^3_6=20\);

The # of ways to arrange the vowels: 3! (or instead of these two steps we could use \(P^3_6)\);

The # of ways to arrange MNFLD = 5!.

Total = 20*3!*5! = 14,400.

Answer: A.
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*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!.

{Desired} = {Total} - {All 3 together} - {Exactly two together} = 8! - 6!*3! - 3*2!*6*5*5! = 14,400.

Hope it's clear.
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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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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 ...

Hope it's clear.
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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....

Hi Bunuel,

I have similar method and getting the same answer but having looked at the solution above, my way of working may not correct. Can you please check

Total no. of possible outcomes for MANIFOLD are: 8!
No. of favourable cases : Total - (Cases in which all 3 vowels are together)

Taking 3 vowels as one unit we can arrange the 6 words in 6! ways and among themselves the vowels will arrange in 3! ways

*M*N*F*L*D*-------> Now the * can be the position of the vowels together so no. of such words will be 6!*3!*6= 25920

So no. of words that can be formed in which no vowels are together : 8!- 25920= 14400

Ans A
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There are two problems with your solution:

1. The same as ratnanideepak made in his approach: in-how-many-ways-can-the-letters-of-the-word-manifold-be-arr-167127.html#p1328143

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}.

Hope it's clear.
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I have done it like this...

Attached a non vowel letter to the right OR left of each vowel and make it one unit. (2 ways)

M AN IF OL D

Now we have 5 entities. we can arrange them in 5! ways.

We have 5 non Vowels and 3 Vowels to attach them to, we can select non-vowels to attach with vowels in 5*4*3 ways.

Total ways: 2*5!*5*4*3 = 14400

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

Answer is A:

My method is different,

consider the following numbers as blank spaces which i will use to denote
1 2 3 4 5 6 7 8

case 1) vowel at 1,3 then third can be at 5,6,7,8 = 4 case
when at 1,4 then = 3 case
1,5= 2case
1,6 = 1 case
total = 10 cases

case 2) when vowel at 2,4 then 3rd can be at 6,7,8 = 3 cases
when at 2,5 = 2cases
2,6 = 1 case

total = 6 cases

case 3) when vowel at 3,5 then 3rd can be at 7,8 = 2 case
when at 3,6 = 1 case
total = 3 cases

case 4) when vowel at 4,6 then 3rd can be at 8 = 1 cases
total = 1 case

total cases = 10+6+3+1 = 20

in each case total number of arrangement = 3! for vowel and 5! for consonants = 5!3!

for 20 cases = 20 x 5! x 3! = 14400 option A

great question ; in hurry i calculated not together later realised questions are separated so 2 vowels can still be together
for consonants ; 5! ways
and vowels AIO can be arranged in b/w in 6*5*4 ways
total 5! *6*5*4 = 120 * 14400 ways
IMO A

Hi Archit3110,

I do not understand the part were we consider that there will be 6 spaces to arrange the 3 vowels after the 5 consonants are arranged. Shouldn't there be only 3 spaces after arranging 5 consonants?

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_

Warm Regards,
Pritishd

Pritishd
Vowels are to be arranged separated.

Below if you see we have 6 spaces for vowels and total vowels are 3 so possible arrangement for vowels in 6 spaces:6*5*4..

_M_N_D_L_F_

Hope this helps



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Hi Archit3110,

My doubt still remains. I can see that there are 6 spaces but my question was that how did you arrive at those 6 spaces? Once we arrange the 5 consonants will we not have only 3 spaces left? Is this some kind of a method were we assume an available space before and after every element?

Warm Regards,
Pritishd

Con first : A55
Vo: 6 slots .. 3 pieces A63

A55 * A63 =14400

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Pritishd

Well this is how arrangement questions are solved we need to make cases as per question and solve considering all possible arrangement...
You can look for similar questions in GMAT club directory and practice similar type questions.



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While I understand the other methods that arrangement questions are solved, it is this type that I am unable to figure out. Let me figure out at my end in that case.

3 vowels AIO and 5 consonants MNFLD to a generalized case: *M*N*F*L*D*, where * can be either empty or filled by any of the three vowels

The number of ways to choose 3 empty slots out of 6 for 3 vowels = 6C3 = 20

The number of ways to arrange the vowels= 3! = 6

The number of ways to arrange MNFLD = 5! = 120

Total = 20*6*120 = 14,400.

FINAL ANSWER IS (A)

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