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.
Griffins said:
8 years ago
What's the meaning of 3P3 as used? Explain me.
Jilsa said:
8 years ago
What if we want to arrange letters of *Machine* such that vowels occupy odd places?
Sourabh said:
8 years ago
0! = 0.
Soham said:
8 years ago
How many no of 7 digits can be formed with digit 1, 2, 3, 4, 3, 2, 1 so that the odd digit always occupy odd place?
Can anyone answer this?
Can anyone answer this?
Aishee said:
7 years ago
The word detail can be arranged with vowels at odd places in 3C1*3C1*2CI*2C1*1C1*1C1 as no of vowels are 3 and consonants are 3 in no.
So 1st place can be arranged in 3C1 ways. Similarly, 2nd place can be arranged in 3C1 ways as there are 3 consonants. Similarly, the 3rd 4th 5th and 6th place can be arranged in 2C1, 2C1, 1C1, 1C1 ways respectively. Total no of ways is their product. This problem can be done with combination also. Am I right?
So 1st place can be arranged in 3C1 ways. Similarly, 2nd place can be arranged in 3C1 ways as there are 3 consonants. Similarly, the 3rd 4th 5th and 6th place can be arranged in 2C1, 2C1, 1C1, 1C1 ways respectively. Total no of ways is their product. This problem can be done with combination also. Am I right?
Ganesh kc said:
7 years ago
I want to know the arrangement of the word "management" in how many ways can we arrange?
Can anyone answer this?
Can anyone answer this?
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!).
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.
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.
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.
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