How many ways can the letters of the word bottles be arranged such that both vowels are at the end

Discussion :: Permutation and Combination - General Questions (Q.No.10)

10.

In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?

[A].	32
[B].	48
[C].	36
[D].	60
[E].	120

Answer: Option C

Explanation:

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

(1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = 3P3 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = 3P3 = 3! = 6.

Total number of ways = (6 x 6) = 36.

Gayathri said: (Apr 30, 2011)
Why we use permutation concept instead of combination.

Manasa said: (Jun 14, 2011)
@Gayathri In this case we are not selecting any letters. Just we are arranging the letters according to given condition.

Kesav said: (Jul 21, 2011)
Can any one solve this problem for me? 1! + 2! + .... + 50!=? a) 3.103510^64 b) 2.102110^65 c) 3.103510^63 d) 3.103510^62

Riddhi said: (Jul 24, 2011)
@kesav:plz give me ans. Is it b?

Vishnu said: (Aug 14, 2011)
It is asked in TCS apti test.

Gautam said: (Dec 28, 2011)

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants. Number of ways of arranging the vowels = nPr= n!/(n-r)= 3/(3-3) =3!/0! ,=3*2*1= 6 {we define 0! = 1} Also, the 3 consonants can be arranged at the remaining 3 positions. Number of ways of arranging the vowels = nPr= n!/(n-r)= 3/(3-3) =3!/0! ,=3*2*1= 6 {we define 0! = 1}.

Total number of ways = (6 x 6) = 36.

Nikhil said: (Jan 30, 2013)
Vowel can even be put at the 7th place, not considered above, which will increase the number of possible arrangements. Please clarify?

Athu said: (Apr 11, 2013)
What is odd positions?

Drushya said: (Apr 20, 2013)
@Athu. Odd position means its like odd no. For example : 1 2 3 4 5 6. In this case odd position is :1 3 5 & even position : 2 4 6.

Jinal said: (Jul 28, 2013)
Vowels can also take position 7 (the last position) ? V C V C V C V. So the answer would be 3!*4C3 = 24. Is this correct?

Vasudha said: (Jul 31, 2013)
Since there is nothing mentioned about repetitions so the no ways could also be 333(i.e., E can come in all three odd places and same goes for the other two vowels A and I as well) for the three odd positions of vowels. And similarly the other three letters can be arranged so the answer should be 33333*3.

Sanjay said: (Sep 1, 2013)
Hi guys. There is 3 odd, 3 even place and the word have 3 vowel and 3consonant. We arrenge 3 vowel in 3 odd place and 3 consonant in 3 rest place such that VCVCVC So total no of way = 3p3*3p3 = 36.

Neha Pant said: (Aug 17, 2014)
But here it is not mentioned that repetition is allowed so according to me the correct way is, 333333 = 729 which is not in option. 33*3 for 3 vowels and other for consonant . Please explain it?

Nee said: (Oct 16, 2014)
Can't we do it in this way like first we find all possible ways of arranging all letters. i.e 6! and then subtract 4!(3!) from 6! to get the answer.

Priya said: (Feb 7, 2015)
Any example of cards combination. Tell me.

PRAKASH S said: (Jul 13, 2015)
I really don't understand why we are doing that. How we are taken 3p3 = 6?

Rohini said: (Jul 24, 2015)
Why write this type 3p3?

Subhadeep said: (Aug 23, 2015)
We can consider the seventh position also. So I feel the answer is wrong.

Raji said: (Aug 24, 2015)
I think we arrange 3 consonants + 1 vowel = 4 (generally 3 vowels but total vowels take as 1 set). So 4! = 12. The three vowels re arrange is 3!=6. So answer is 72. Is there any wrong, please tell me.

Sourav Sarkar said: (Sep 1, 2015)
Well, there are four odd positions according to me. 1234567. 1, 3, 5, 7 are the odd positions. 3 vowels have to be selected to fit in any of the three position out of four. This can be done in 4C3 ways. The rest remains the same. Answer- 4C3x3!x3!

Shivam said: (Sep 3, 2015)
There are 4 possible position for the vowel so why 4C3 is not included in the solution for selecting 3 position out of 4 for vowels?

Balakrishna said: (Sep 4, 2015)
Here they didn't mention about repetition allowed or not if we do this problem with repetition allowed the answer may be not 36.

Devdp said: (Nov 24, 2015)
When 1st try myself consider x vowel and y consonants it can be possibility of position (1) xyxyxy and (2) yxyxyx. As per permutation for x 3p3 = 3! = 6. Permutation for why 3y3 = 3! = 6. And as shown as above 2 different position. Total arrangement = 662 = 72 (Ans). But when I shows option I confused. There is no 72.

Devdp said: (Nov 25, 2015)
In 3rd line permutation for why 3p3 = 3! = 6. This small mistake done bye me in writing but answer I got is 72. Reply anyone.

Divyansh said: (Nov 27, 2015)
Brother only odd position so answer is 36.

Dev said: (Dec 22, 2015)
It can be done by 6x5x4x3x2 divide by 2 and again divide by 10 = 36.

Pinky Kumari said: (Aug 10, 2016)
The answer will be 36. It is correct. Because there is 3 odd place for vowel and 3 for a consonant.

Aditya said: (Aug 12, 2016)
How can we solve the same question taking 'FATHER' as the given word?

M223 said: (Aug 21, 2016)
How many ways can the word DETAIL be rearranged such that no two vowels are placed adjacently? Please tell me the answer.

Aman said: (Sep 1, 2016)

Why making this question so complicated? See we have the word "DETAIL" which is having the total number of alphabets that is "6". In which "3" is vowels and "3" is consonants and we have to arrange the vowels at odd places right so what are the positions we have (1) 2 (3) 4 (5) 6 [number inside the braces indicates the odd places]. So we have 3 places and we need to arrange the 3 vowels we do it by combination that is 3c3=1 and the no of vowels is 3 so their permutation is 3! so that the similar permutation of consonants is having the same 3! because we have 3 consonants as well so finally the answer is => 3C3 * 3! * 3! => 1* 6 * 6.

=> 36.

Aman said: (Sep 1, 2016)

@Aditya. I have the solution for your question you have the word "FATHER" that is nothing but 6 letters in which 2 vowels and 4 consonants. So we have to arrange the vowels at odd places as we know that out of 6 words 3 places are odd places, so the arrangement of 2 vowels at 3 places means 3C2 right. So we arranged that vowel at odd places now we have to take the permutation of that vowels as 2! and the permutation of the consonants that is nothing but 4! so finally we have the solution as :- => 3C2 * 2! * 4! => 3 * 2 * 24.

=> 144.

Kat@93 said: (Sep 12, 2016)
@Devdp. I too agree with you. If I name the places from left-hand side i.e. 6-5-4-3-2-1 or from right-hand side 1-2-3-4-5-6. I will get 2 types of arrangement and since the question is regarding permutation both way of arrangement matters, so answer we get is 72.

Gopika S S said: (Sep 28, 2016)

There are 6 positions out of which 1, 3 and 5 are for the three vowels. There are 3 ways for arranging in position 5 and remaining 2 ways for position 3 and 1 way for position 1. So for vowels, it is 3 * 2 * 1 = 6. For remaining three consonants which are placed in even position, there are 3 ways in position 6 and remaining 2 ways in position 4 and one way in position 2. So for consonants, it is 3 * 2 * 1 = 6.

So a total number of ways = 6 * 6 = 36.

Tommy said: (Oct 6, 2016)

It is 36. Assuming we represent the filings as D*T**L. So that d (*) the possible fillings for the vowels (E A I). Then "E" can go into the arrangement in 3ways I. E either "E" takes d first, second, or the third filling space ie shown by the (*) sign. So after E as occupying one space, then we are left with just spaces left which are to be filled by "A" and "I". In which "A" can go into it in 2ways. Remaining one space to be occupied by "I" in just 1way.

So arrangement of vowel is 3P3 = 3! = 3 * 2 * 1 = 6. So we have 3 consonant (D T L) left we can permute (arrange) them in 3! ways = 3 * 2 * 1 = 6 ways. So total arrangement = 6 * 6 = 36 ways.

Bonface said: (Nov 20, 2016)
Why we also considering the positions of the consonants?

Sanif said: (Nov 26, 2016)
In how many ways can the letters of the word SYSTEM be arranged so that E is always at the third position?

Babar said: (Nov 29, 2016)
How there are three vowels in the word vowels? logically we cannot consider the word "vowels" as a vowel. Can anyone explain, please?

Sumit said: (Dec 4, 2016)
In how many ways can the letter of word failure be arranged si that the consonants may occupy only odd position.

Shekhar said: (Mar 6, 2017)
In how many ways can the letters of the word CIVIL be rearranged? Can anyone explain? Please.

Sai Reddy said: (Mar 23, 2017)
@Shekhar. Civil = 5!/2! = 60.

Isidore said: (Apr 21, 2017)
Please give me the answer. What if the question is, In how many different ways can the letters of the word "PUNCTUATE" be arranged in such a way that the vowels occupy only the even positions? Observe the vowels: UUAE (here 'U' occurs twice).

Omhari said: (May 21, 2017)
In how many ways we can arrange the alphabets of the word 'ARRANGE' such that 2a and 2r do not come together? Please solve this.

Griffins said: (May 24, 2017)
What's the meaning of 3P3 as used? Explain me.

Jilsa said: (Aug 13, 2017)
What if we want to arrange letters of Machine such that vowels occupy odd places?

Sourabh said: (Aug 25, 2017)
0! = 0.

Soham said: (Aug 25, 2017)
How many no of 7 digits can be formed with digit 1, 2, 3, 4, 3, 2, 1 so that the odd digit always occupy odd place? Can anyone answer this?

Aishee said: (Jan 31, 2018)

The word detail can be arranged with vowels at odd places in 3C1*3C1*2CI*2C1*1C1*1C1 as no of vowels are 3 and consonants are 3 in no.

So 1st place can be arranged in 3C1 ways. Similarly, 2nd place can be arranged in 3C1 ways as there are 3 consonants. Similarly, the 3rd 4th 5th and 6th place can be arranged in 2C1, 2C1, 1C1, 1C1 ways respectively. Total no of ways is their product. This problem can be done with combination also. Am I right?

Ganesh Kc said: (Feb 26, 2018)

I want to know the arrangement of the word "management" in how many ways can we arrange?

Can anyone answer this?

Daniel said: (Mar 2, 2018)
With using all 6 letters we can arrange - 6! If 3 vowels are not odd - 4!3!. so if we decrease these to answers we can get the answer for this question = 6!-(4!3!).

Uthaya said: (Mar 8, 2018)
@GANESH. Management has 10 letters, so 10! = 3628800. the letters m,a,n,e has repetition. So 10!/2!2!2!2! = 3628800/(222*2) = 226800.

Usha said: (Aug 24, 2018)
Then what is the use of vowels arranged in odd position? I think it's wrong.

Navi said: (Nov 12, 2018)
Simple: (DTL) * (EAI). (321) * (321). (6)*(6) = 36.

Sahana said: (Apr 7, 2019)
We can arrange consonants in 3! ways in even position and 4 odd places can be occupied by 3 vowels in 4P3 ways. So these can be arranged in 3!*4P3 =144 ways. Am I right?

Sayani said: (Jun 23, 2019)
If we can set the vowels in odd positions, we can get one combination like; ED AT IL. Like that; So always one consonant comes along with a vowel. So total combined words= 3! And 3 vowels can arrange themselves in 3! -> Number Of ways. So, total arrangement is= 3! *3!. =36.

PREM said: (Aug 16, 2019)
SIMPLE TRICK: There are 6 Letters 3 V and 3 C. 6blank positions_ _ _ _ _ _. Which can be filled like this VCVCVC. 3V3C2V2C1V1C = 3 * 3 * 2 * 2 * 1 * 1 = 36.

Omshiva said: (Sep 5, 2019)
@Keshav. By Using the Gauss method. The correct answer is A.

Dinesh S said: (Oct 10, 2019)
DETAIL = vowels ( E A I) = 3! No.of possible is(DETAIL) _ _ _ _ _ _ 1 2 3 4 5 6 letter so conditions is odd position only 1,3,5,place only possible....=3! Then 3! x 3! = 36.

Divya said: (Jun 23, 2020)
@Devdp. As per the question, Vowels can occupy only the odd position. But you have placed the vowel in even position. Explain clearly.