Aptitude - Permutation and Combination - Discussion
Discussion Forum : 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?
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, 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.
Discussion:
59 comments Page 1 of 6.
Avinash aswal said:
4 years ago
Here question says vowels occupy only the odd position, then what about consonants? Please clarify this.
(3)
Divya said:
5 years ago
@Devdp.
As per the question, Vowels can occupy only the odd position. But you have placed the vowel in even position. Explain clearly.
As per the question, Vowels can occupy only the odd position. But you have placed the vowel in even position. Explain clearly.
(5)
Dinesh S said:
6 years ago
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.
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.
(4)
Omshiva said:
6 years ago
@Keshav.
By Using the Gauss method. The correct answer is A.
By Using the Gauss method. The correct answer is A.
(6)
PREM said:
6 years ago
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.
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.
(22)
Sayani said:
6 years ago
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.
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.
Sahana said:
6 years ago
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?
So these can be arranged in 3!*4P3 =144 ways.
Am I right?
(1)
Navi said:
7 years ago
Simple:
(DTL) * (EAI).
(3*2*1) * (3*2*1).
(6)*(6) = 36.
(DTL) * (EAI).
(3*2*1) * (3*2*1).
(6)*(6) = 36.
(2)
Usha said:
7 years ago
Then what is the use of vowels arranged in odd position?
I think it's wrong.
I think it's wrong.
Uthaya said:
7 years ago
@GANESH.
Management has 10 letters, so 10! = 3628800.
the letters m,a,n,e has repetition. So 10!/2!2!2!2! = 3628800/(2*2*2*2) = 226800.
Management has 10 letters, so 10! = 3628800.
the letters m,a,n,e has repetition. So 10!/2!2!2!2! = 3628800/(2*2*2*2) = 226800.
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