Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 4)
4.
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Answer: Option
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
| = (7C3 x 4C2) | |||||||||
|
|||||||||
| = 210. |
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
| Number of ways of arranging 5 letters among themselves |
= 5! |
| = 5 x 4 x 3 x 2 x 1 | |
| = 120. |
Required number of ways = (210 x 120) = 25200.
Video Explanation: https://youtu.be/dm-8T8Si5lg
Discussion:
65 comments Page 3 of 7.
Sandhya said:
9 years ago
3 out of 7 consonants,
7 * 6 * 5/3 * 2 * 1 = 210/6
2 out of 4 vowels 4 * 3/2 * 1 = 12/2 = 6,
210/6 * 6 = 210.
Formation of 3 Consonants and 2 vowels 3 + 2 = 5.
5 * 4 * 3 * 2 * 1 = 120,
210 * 120 = 25200.
7 * 6 * 5/3 * 2 * 1 = 210/6
2 out of 4 vowels 4 * 3/2 * 1 = 12/2 = 6,
210/6 * 6 = 210.
Formation of 3 Consonants and 2 vowels 3 + 2 = 5.
5 * 4 * 3 * 2 * 1 = 120,
210 * 120 = 25200.
Mohan said:
9 years ago
It's.
VCVCC
VCVCV
VCCVC
VCCCV
CVVCC
CCVVC
CCVVC
CCVCV
CCcvv
VCVCC
VCVCV
VCCVC
VCCCV
CVVCC
CCVVC
CCVVC
CCVCV
CCcvv
Aditya said:
9 years ago
It's simple,
Select 3 consonants out of B,C,D,F,G,H,J (assume these 7 consonants) in 7P3 ways.
So, [ B C D ] is one selection out of 7P3 ways right?
but, [ C B D] , [ C D B ] , [ B D C ] , [ D C B ] , [ D B C ] , all these contain same consonants.
This is why you use the combination, to form groups of 3 consonants out of 7 consonants.
Same goes for vowels : 4P2 ways to get groups of 2 vowels out of 4 vowels.
total groups of consonants and vowels : 7P3 * 4P2.
Now, we need to form words of 5 letters consisting of vowels and consonants.
How many ways? In 5! ways.
Thus, 7P3 * 4P2 * 5! = 25200 ways.
Select 3 consonants out of B,C,D,F,G,H,J (assume these 7 consonants) in 7P3 ways.
So, [ B C D ] is one selection out of 7P3 ways right?
but, [ C B D] , [ C D B ] , [ B D C ] , [ D C B ] , [ D B C ] , all these contain same consonants.
This is why you use the combination, to form groups of 3 consonants out of 7 consonants.
Same goes for vowels : 4P2 ways to get groups of 2 vowels out of 4 vowels.
total groups of consonants and vowels : 7P3 * 4P2.
Now, we need to form words of 5 letters consisting of vowels and consonants.
How many ways? In 5! ways.
Thus, 7P3 * 4P2 * 5! = 25200 ways.
Hehe said:
9 years ago
I don't think that it's supposed to be multiplied by 5! since it doesn't ask for the arrangement of the letters.
Neeraj Tailor said:
9 years ago
How many numbers greater than 5000 can be formed with the digits, 3, 5, 7, 8, 9 no digit being repeated?
In this question we won't multiply instead of adding?
In this question we won't multiply instead of adding?
Prasant said:
1 decade ago
Finally multiplied with 5!, because, each word is consisted of 5 distinct letters. Had it been persons in place on letters, then 5! multiplication would not have been applied.
Benni said:
1 decade ago
How can we find permutation and combination problem ?
Rizwana said:
1 decade ago
Hi,
I always find problem in using C and P in formula I could not distinguish when to use nCr and when to use nPr please help in finding difference.
I always find problem in using C and P in formula I could not distinguish when to use nCr and when to use nPr please help in finding difference.
Atul Moundekar said:
10 years ago
This question is only based on a combination, they never told to find the arrangements. So the answer is 210.
Litu said:
1 decade ago
Hi,
I do agree with Sachin Jain as it is a permutation query. We can arrange it in any way using using 3 consonants out of 7 & 2 vowels out of 4.
Ex: If there is bcdfghj then we can do it bcd n also cbd so it is permutation type not combination so the answer should be 7P3 * 4P2 = 2520
But again we have to arrange these 2 groups together. So I'm not getting form this point.
Can anyone clear the same?
I do agree with Sachin Jain as it is a permutation query. We can arrange it in any way using using 3 consonants out of 7 & 2 vowels out of 4.
Ex: If there is bcdfghj then we can do it bcd n also cbd so it is permutation type not combination so the answer should be 7P3 * 4P2 = 2520
But again we have to arrange these 2 groups together. So I'm not getting form this point.
Can anyone clear the same?
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