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 1 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.
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.
Anushq said:
1 decade ago
Hi all,
I have a doubt in questions regarding vowels and consonants. If we arrange all the vowels in one word, it can be in either consonant + (sum of vowels) or (sum of vowels) + consonant. Why all the answers are not considering the second option.
Because if we consider both the case then We need to multiply all the answers * 2. Please let me know if I am wrong.
I have a doubt in questions regarding vowels and consonants. If we arrange all the vowels in one word, it can be in either consonant + (sum of vowels) or (sum of vowels) + consonant. Why all the answers are not considering the second option.
Because if we consider both the case then We need to multiply all the answers * 2. Please let me know if I am wrong.
Ayesha said:
6 years ago
OAI are 3 vowels should be considered as one letter (OAI) and the 3 vowels among themselves can be arranged in 3! ways ie.,6ways.
Now the remains consonants are PCTL are 4 letters plus we should consider the 3 vowels as one letter (4+1)=5! ways {5!=120}.
The required no.of ways = 120 * 6 =720 ways.
Now the remains consonants are PCTL are 4 letters plus we should consider the 3 vowels as one letter (4+1)=5! ways {5!=120}.
The required no.of ways = 120 * 6 =720 ways.
(3)
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/
Sandeepkumar said:
1 decade ago
(1) In letter OPTICAL there seven letters out of which 3 vowels and 4 consonants.
(2) If vowels are together they can be arranged among themselves in 3! ways.
(3) Now considering them as single term (1+4=5) they can be arranged among themselves in 5!ways
Total no.of ways=5!*3!=120*6=720
(2) If vowels are together they can be arranged among themselves in 3! ways.
(3) Now considering them as single term (1+4=5) they can be arranged among themselves in 5!ways
Total no.of ways=5!*3!=120*6=720
Praveen said:
1 decade ago
There are 4 letters in this, so we can arrange those 4 letters in 4!ways and not in 5! ways..
Now, 4 letters can be arranged in 4!=24
The vowels (OIA) can be arranged among themselves in 3!=6 ways
Required number of ways =(24*6)=144
Now, 4 letters can be arranged in 4!=24
The vowels (OIA) can be arranged among themselves in 3!=6 ways
Required number of ways =(24*6)=144
Navya said:
1 decade ago
Ibidun: 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.
Sol: 5c2*9c3 ways.
Sol: 5c2*9c3 ways.
Sugumaran said:
1 decade ago
@Jiyas.
Convert 30 min into sec, (i.e) 30*60=1800 sec.
Find the LCM of 2, 4, 6, 8, 10, 12=120.
That is every 120 seconds all the 6 bells are ringing together.
For 1800 sec=1800/120= 15 times.
Convert 30 min into sec, (i.e) 30*60=1800 sec.
Find the LCM of 2, 4, 6, 8, 10, 12=120.
That is every 120 seconds all the 6 bells are ringing together.
For 1800 sec=1800/120= 15 times.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers