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 5 of 7.
Achu said:
1 decade ago
What's the need to take 5 here? They didn't mention any groupings here. Then what's the purpose to group?
Math Wiz said:
1 decade ago
NO. It's not because of the letters (and not person) we should multiply with 5!, but it's about what the question said; how many words can be FORMED? that's we call PERMUTATION. At first we might have thought it's just COMBINATION because we didn't notice the last phrase of the question said how many words can be FORMED. So it's considered as both PERMUTATION and COMBINATION.
Shub said:
1 decade ago
I don't know why all are confusing as permutation and combination.
I think this question"how many words can be FORMED " is clearly combination.
I think this question"how many words can be FORMED " is clearly combination.
Harsha said:
1 decade ago
When I saw this question I was like it will be permutation since arrangement of words is important and I did 7P3*4P2(just like other sums).
Why does the method differ here !
Why does the method differ here !
Shams said:
1 decade ago
@Harsha.
Here the asking question is group of words (use combination) not to asking about the order of words so don't use the permutation. OK.
Here the asking question is group of words (use combination) not to asking about the order of words so don't use the permutation. OK.
Harsha vardhan said:
1 decade ago
Here we should use permutations we need to use combinations and just selecting the consonants and vowels from the given 7 consonants and 4 vowels, its not necessary to multiply with 120 there so according to me, its 210 just selecting those consonants and vowels from the given one not more than that.
Akhila said:
1 decade ago
How can we find the given problem is from permutation or combination?
Kaustav said:
1 decade ago
Combination 3 consonant and 2 vowels is still left. By adding forms 5! and then rest to find the total.
Uday sai ranjan said:
1 decade ago
We can do this problem with permutation.
Let me explain.
We got 2520 by permutation but its not complete. It is only the permutation of two sets of alphabets. So what we have missed out is combining both the sets and the number of ways of combining the two sets (3c and 2v) is 10. So finally we get 25200 as the answer.
They are:
vvccc
vcvcc
vccvc
vcccv
cvvcc
cvcvc
cvccv
ccvvc
ccvcv
cccvv
Let me explain.
We got 2520 by permutation but its not complete. It is only the permutation of two sets of alphabets. So what we have missed out is combining both the sets and the number of ways of combining the two sets (3c and 2v) is 10. So finally we get 25200 as the answer.
They are:
vvccc
vcvcc
vccvc
vcccv
cvvcc
cvcvc
cvccv
ccvvc
ccvcv
cccvv
Jyoti bala said:
1 decade ago
When we apply permutation and combination? please tell me one.
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