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Aptitude - Permutation and Combination - Discussion

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"Two things are infinite: the universe and human stupidity; and I'm not sure about the universe."
- Albert Einstein
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

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: (Sat, Apr 30, 2011 03:32:24 AM)    
 
Why we use permutation concept instead of combination.

Manasa said: (Tue, Jun 14, 2011 07:25:39 AM)    
 
@Gayathri

In this case we are not selecting any letters. Just we are arranging the letters according to given condition.

Kesav said: (Thu, Jul 21, 2011 07:57:15 PM)    
 
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: (Sun, Jul 24, 2011 12:56:19 AM)    
 
@kesav:plz give me ans. Is it b?

Vishnu said: (Sun, Aug 14, 2011 08:57:33 PM)    
 
It is asked in TCS apti test.

Gautam said: (Wed, Dec 28, 2011 05:59:07 PM)    
 
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: (Wed, Jan 30, 2013 09:48:32 AM)    
 
Vowel can even be put at the 7th place, not considered above, which will increase the number of possible arrangements. Please clarify?

Athu said: (Thu, Apr 11, 2013 02:49:17 AM)    
 
What is odd positions?

Drushya said: (Sat, Apr 20, 2013 05:30:13 PM)    
 
@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: (Sun, Jul 28, 2013 12:59:51 PM)    
 
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: (Wed, Jul 31, 2013 11:31:50 PM)    
 
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: (Sun, Sep 1, 2013 07:54:50 PM)    
 
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: (Sun, Aug 17, 2014 03:26:51 PM)    
 
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: (Thu, Oct 16, 2014 02:27:05 AM)    
 
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: (Sat, Feb 7, 2015 10:24:10 PM)    
 
Any example of cards combination. Tell me.

Prakash S said: (Mon, Jul 13, 2015 03:20:12 PM)    
 
I really don't understand why we are doing that.

How we are taken 3p3 = 6?

Rohini said: (Fri, Jul 24, 2015 12:21:04 PM)    
 
Why write this type 3p3?

Subhadeep said: (Sun, Aug 23, 2015 08:17:47 PM)    
 
We can consider the seventh position also. So I feel the answer is wrong.

Raji said: (Mon, Aug 24, 2015 12:40:34 PM)    
 
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: (Tue, Sep 1, 2015 03:07:06 PM)    
 
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: (Thu, Sep 3, 2015 10:32:16 AM)    
 
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: (Fri, Sep 4, 2015 12:12:29 PM)    
 
Here they didn't mention about repetition allowed or not if we do this problem with repetition allowed the answer may be not 36.

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