Aptitude - Permutation and Combination - Discussion

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) 3.1035*10^64
b) 2.1021*10^65
c) 3.1035*10^63
d) 3.1035*10^62

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 3*3*3(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 3*3*3*3*3*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,

3*3*3*3*3*3 = 729 which is not in option.
3*3*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 = 6*6*2 = 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.

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