Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 14)
14.
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
Answer: Option
Explanation:
The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
Discussion:
37 comments Page 2 of 4.
Simmie said:
9 years ago
@Sonam.
If non-vowels come together then,
Vowels OIA considering each as different units So its 3 units,
Non-vowels, PTCL as 1 units.
Its 4!=24 because they need to be arranged within themselves.
The vowels and non-vowels together they make 4 units..4!=24(its like consider "PTCL" as "x" ..now arranging them with vowels would be OIAX,OAIX,AIOX,IOAX,IAOX......i.e....4!)
So now,
Vowels and non-vowels = 4! = 24,
Non-vowels = 4! = 24,
24 * 24 = 576.
If non-vowels come together then,
Vowels OIA considering each as different units So its 3 units,
Non-vowels, PTCL as 1 units.
Its 4!=24 because they need to be arranged within themselves.
The vowels and non-vowels together they make 4 units..4!=24(its like consider "PTCL" as "x" ..now arranging them with vowels would be OIAX,OAIX,AIOX,IOAX,IAOX......i.e....4!)
So now,
Vowels and non-vowels = 4! = 24,
Non-vowels = 4! = 24,
24 * 24 = 576.
Mahesh Babu said:
9 years ago
@ Simmie.
You did a mistake the correct procedure is;
Since no vowels come together, first arrange the 4 consonants in 4! ways.
Now, a vowel can be placed at the beginning of the consonants or at the end of the consonants or in between every two consonants.
_C_C_C_C_
They are 5 places to fill 3 vowels,
In these 5 places we can arrange the 3 vowels in 5!/2! ways(5p3 ways),
The number of words in which no vowels come together is,
4!*5!/2!=1440.
You did a mistake the correct procedure is;
Since no vowels come together, first arrange the 4 consonants in 4! ways.
Now, a vowel can be placed at the beginning of the consonants or at the end of the consonants or in between every two consonants.
_C_C_C_C_
They are 5 places to fill 3 vowels,
In these 5 places we can arrange the 3 vowels in 5!/2! ways(5p3 ways),
The number of words in which no vowels come together is,
4!*5!/2!=1440.
Sree vidya said:
9 years ago
When to use permutation & when to use a combination? Please, someone, explain it.
Thaslim said:
9 years ago
When order takes important place then it is a permutation.
In Combination, order doesn't play an important role.
In Combination, order doesn't play an important role.
BadShah KinG said:
8 years ago
How many five different letter words can be formed out of the word "LOGARITHMS" ?
Can anyone solve this?
Can anyone solve this?
Mukesh said:
1 decade ago
Can we use the same method in question 13?
Gaurav said:
8 years ago
Short trick of that question is;
Total letters=7!,
Total vowels=3!(or 1group),
7!-3!=4!,
4!+1!(group of vowel)=5!,
Num of ways=5!*3!=720 is the answer.
Total letters=7!,
Total vowels=3!(or 1group),
7!-3!=4!,
4!+1!(group of vowel)=5!,
Num of ways=5!*3!=720 is the answer.
Karan kumar said:
8 years ago
Vowels come together so there are 5!
And 3!.
120 * 6 = 720.
And 3!.
120 * 6 = 720.
Shashank Bajpayee said:
2 weeks ago
It can also be made by using a shortcut formula.
Like if we have asked about vowels occurring together and there are no repeated vowels, then we can use this formula.
2!(n-1)!, 3! (n-2)!
Similarly,
Now here there are 3 vowels.
So, 3!(n-2)!.
3!(7-2)! = 120 => answer.
Like if we have asked about vowels occurring together and there are no repeated vowels, then we can use this formula.
2!(n-1)!, 3! (n-2)!
Similarly,
Now here there are 3 vowels.
So, 3!(n-2)!.
3!(7-2)! = 120 => answer.
Santosh said:
1 decade ago
(1) In 7 letter word "OPTICAL" has 3 vowels. They are 'O','I' & 'A'. We can arrange them in 3! ways and the remaining 4 letters arranged in 4! ways.
(2) If we combine OIA, it occurs in 5 positions of entire lenght of the word.
(3) So, we have 5*3!*4! ways to arrange the given word/
(2) If we combine OIA, it occurs in 5 positions of entire lenght of the word.
(3) So, we have 5*3!*4! ways to arrange the given word/
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