Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 3)
3.
In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
Answer: Option
Explanation:
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
| Number of ways arranging these letters = | 7! | = 2520. |
| 2! |
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
| in | 5! | = 20 ways. |
| 3! |
Required number of ways = (2520 x 20) = 50400.
Video Explanation: https://youtu.be/o3fwMoB0duw
Discussion:
60 comments Page 2 of 6.
BHAGYESH said:
6 years ago
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
The number of ways of arranging these letters = 7!/2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
In 5!/3! = 20 ways.
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
The number of ways of arranging these letters = 7!/2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
In 5!/3! = 20 ways.
(7)
Nishtha sharma said:
1 decade ago
Here we can simply separate out the common alphabet.
in CORPORATION , o is 3 times and r is two times, putting together we have,OOORRCPATIN . now the total unique alphabets are 7 and thus the answer is 7%=7*6*5*4*3*2*1= 50400 . the common alphabets dun need any permutation.
in CORPORATION , o is 3 times and r is two times, putting together we have,OOORRCPATIN . now the total unique alphabets are 7 and thus the answer is 7%=7*6*5*4*3*2*1= 50400 . the common alphabets dun need any permutation.
Veera manikantha said:
7 years ago
In CORPORATION their vowels have to be together, so it becomes CRPRTN (OOAIO).
In this, a number of ways is 7!÷2! since R is repeated 2 times and in the vowels, there are 5!÷3! A number of ways. So total number of ways is (7!÷2!) * (5!÷3!)=50400.
In this, a number of ways is 7!÷2! since R is repeated 2 times and in the vowels, there are 5!÷3! A number of ways. So total number of ways is (7!÷2!) * (5!÷3!)=50400.
(4)
Naveen said:
8 years ago
@Gbenga.
Total there are 5members while 2are always together so considering 2members as -unit arrangement can be done in 4! Ways and they themselves i.e the principle and wife can be arranged in 2! ways so total no of ways will be 4!*2! = 48ways.
Total there are 5members while 2are always together so considering 2members as -unit arrangement can be done in 4! Ways and they themselves i.e the principle and wife can be arranged in 2! ways so total no of ways will be 4!*2! = 48ways.
Bob said:
9 years ago
Where in question 3 does it indicate that only one vowel and only one consonant from the alphabet are permitted? Since this question appears to duplicate the wording of question 2, why isn't it solved in a like manner ? What am I missing here ?
Sirius said:
1 decade ago
There is some flaw on the questions, it should be stated "all vowels". If not, it can mean 3 vowel together and the other 2 vowel group together but this 2 group can separate differently. Example O O_ _ _ A I O _ _ _ .
Gbenga Felix said:
8 years ago
A school principal and his wife, as well as three other tutors are to be seated in a row so that the principal and his wife are next to each other. Find the total number of ways this can be done.
Please give me the answer.
Please give me the answer.
Venkata dinesh said:
9 years ago
The give word consistent of 5 vowels and remaining 6 letters + 1 group of vowels is 7!. So to get answer we must do 5! * 7!.
Is this correct? I am getting wrong answer.
Please Help me.
Is this correct? I am getting wrong answer.
Please Help me.
Ambrose said:
10 years ago
Given a set [ABCDEFGHIJKLMNOPQW]. In how many ways can we form a non repeated sub-set of 4-letter with letters (AB) being constant in every sub-set. Example [ABCE], [ABDC].
Abhishek said:
1 decade ago
Why we divide the vowel from all letters?and 1 more why we also calculate the vowel group also because we consider vowel as 1 and we had adjusted as a 7!. Then why again?
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