Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 2)
2.
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
Answer: Option
Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
Video Explanation: https://youtu.be/WCEF3iW3H2c
Discussion:
97 comments Page 5 of 10.
Jomson joy said:
9 years ago
In PACKET there are 2 vowels. Vowels came 2gether means 4! * 2! = 240.
Total words formed =6! (because of total letters) = 720.
Therefore 720 - 240 = 480.
Total words formed =6! (because of total letters) = 720.
Therefore 720 - 240 = 480.
KIRUPA RANI D said:
9 years ago
Can you give solution for this problem? How many words can be formed from the letters of the word 'PACKET', so that the vowels are never together?
Technothelon said:
9 years ago
@Brian.
There aren't such kind of ways. Because arrangement is not taken into consideration when we do combinations.
There aren't such kind of ways. Because arrangement is not taken into consideration when we do combinations.
Brian said:
9 years ago
Correct me if I'm wrong but,
There are 7 letters:
L E A D I N G
3 vowels and 5 non-vowels.
I got the part where we need to permutate 3 and 5, resulting with 3!*5!. But aren't there also several other positions that these vowels could be positioned?
For Example:
AIELDNG is one factor, another factor is LAIEDNG, another factor is LDAIENG.
As you can see, the positions of the 3 vowels with each other are the same, and the order of non-vowels in the word is also the same, but I only changed the position of the starting point for the vowel permutations, starting from the first position, to the second, to the third. So, I believe that the answer should be 3!*5!*5. Since there are 5 different ways you could represent the same order of vowels in different positions with the same order of non-vowels.
Please do correct me if I'm wrong or if I misunderstood the question
There are 7 letters:
L E A D I N G
3 vowels and 5 non-vowels.
I got the part where we need to permutate 3 and 5, resulting with 3!*5!. But aren't there also several other positions that these vowels could be positioned?
For Example:
AIELDNG is one factor, another factor is LAIEDNG, another factor is LDAIENG.
As you can see, the positions of the 3 vowels with each other are the same, and the order of non-vowels in the word is also the same, but I only changed the position of the starting point for the vowel permutations, starting from the first position, to the second, to the third. So, I believe that the answer should be 3!*5!*5. Since there are 5 different ways you could represent the same order of vowels in different positions with the same order of non-vowels.
Please do correct me if I'm wrong or if I misunderstood the question
Siva said:
9 years ago
Other than vowels there are only 4 letter then how it s possible to get 5!.
Peter said:
9 years ago
If at least two vowels always together then what will be the answer for LEADING ?
Eswaru said:
10 years ago
Hey dude we have to form 7 letter words from LEADING that means we need to use all the letters at a time. So no repetition are allowed.
Mukesh said:
10 years ago
Please anyone help me why repetition is not considered in the current problem?
Yonatan said:
10 years ago
I can't understand the answer.
RAJA said:
10 years ago
Hai @Soumya.
S its 1 letter only. Hope you understand.
S its 1 letter only. Hope you understand.
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