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.
PREM said:
7 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.
(23)
Omshiva said:
7 years ago
@Keshav.
By Using the Gauss method. The correct answer is A.
By Using the Gauss method. The correct answer is A.
(6)
Divya said:
6 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:
7 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.
(5)
Avinash aswal said:
5 years ago
Here question says vowels occupy only the odd position, then what about consonants? Please clarify this.
(3)
Sahana said:
7 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?
(2)
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)
Sanjay said:
1 decade ago
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.
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.
(1)
Isidore said:
9 years ago
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).
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).
Kat@93 said:
10 years ago
@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.
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.
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