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:
36 comments Page 2 of 4.
Simmie said:
8 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:
8 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:
8 years ago
When to use permutation & when to use a combination? Please, someone, explain it.
Thaslim said:
8 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?
Manog said:
7 years ago
@Badshah King.
It is 10p5.
It is 10p5.
Gaurav said:
7 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:
7 years ago
Vowels come together so there are 5!
And 3!.
120 * 6 = 720.
And 3!.
120 * 6 = 720.
Ibidun said:
1 decade ago
Please can some please help in solving this with explanation.
(1) A committee of 4 men and 2 women is selected from 10 men and 5 women. If 2 of the men are feuding and will not serve on the committee together, in how many ways can the committee be selected.
(2) In how many ways can a football team be selected from 15 players? IN how many ways if 6 particular players must be included in the team.
(1) A committee of 4 men and 2 women is selected from 10 men and 5 women. If 2 of the men are feuding and will not serve on the committee together, in how many ways can the committee be selected.
(2) In how many ways can a football team be selected from 15 players? IN how many ways if 6 particular players must be included in the team.
Chandu said:
1 decade ago
@Kishore
5!=5*4*3*2*1=120
3!=3*2*1=6
5!=5*4*3*2*1=120
3!=3*2*1=6
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