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 3 of 6.
Shivam said:
10 years ago
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:
10 years ago
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:
10 years ago
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.
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:
10 years ago
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.
This small mistake done bye me in writing but answer I got is 72.
Reply anyone.
Divyansh said:
10 years ago
Brother only odd position so answer is 36.
Dev said:
10 years ago
It can be done by 6x5x4x3x2 divide by 2 and again divide by 10 = 36.
Pinky kumari said:
9 years ago
The answer will be 36. It is correct. Because there is 3 odd place for vowel and 3 for a consonant.
Aditya said:
9 years ago
How can we solve the same question taking 'FATHER' as the given word?
M223 said:
9 years ago
How many ways can the word DETAIL be rearranged such that no two vowels are placed adjacently?
Please tell me the answer.
Please tell me the answer.
Aman said:
9 years ago
Why making this question so complicated?
See we have the word "DETAIL" which is having the total number of alphabets that is "6".
In which "3" is vowels and "3" is consonants and we have to arrange the vowels at odd places right so what are the positions we have (1) 2 (3) 4 (5) 6 [number inside the braces indicates the odd places].
So we have 3 places and we need to arrange the 3 vowels we do it by combination that is 3c3=1 and the no of vowels is 3 so their permutation is 3! so that the similar permutation of consonants is having the same 3! because we have 3 consonants as well so finally the answer is
=> 3C3 * 3! * 3!
=> 1* 6 * 6.
=> 36.
See we have the word "DETAIL" which is having the total number of alphabets that is "6".
In which "3" is vowels and "3" is consonants and we have to arrange the vowels at odd places right so what are the positions we have (1) 2 (3) 4 (5) 6 [number inside the braces indicates the odd places].
So we have 3 places and we need to arrange the 3 vowels we do it by combination that is 3c3=1 and the no of vowels is 3 so their permutation is 3! so that the similar permutation of consonants is having the same 3! because we have 3 consonants as well so finally the answer is
=> 3C3 * 3! * 3!
=> 1* 6 * 6.
=> 36.
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