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 2 of 6.
Sourav Sarkar said:
10 years ago
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!
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!
Isidore said:
8 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).
Neha pant said:
1 decade ago
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?
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?
Raji said:
1 decade ago
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.
So 4! = 12.
The three vowels re arrange is 3!=6. So answer is 72.
Is there any wrong, please tell me.
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.
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)
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)
Daniel said:
7 years ago
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!).
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!).
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)
Nee said:
1 decade ago
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.
i.e 6! and then subtract 4!(3!) from 6! to get the answer.
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