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 2 of 7.
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
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.
Munjal said:
1 decade ago
Difficult question :
First of all it is not the question of combination, it is of permutation (assumed : letters without repetition).
Reason : suppose there are two letters I want to arrange le t say they are t and p, so tp and pt conveys the different words.
So order matters here, so we use 7P3*4P2 instead of 7C3*4C2.
So the answer is 5!*7P3*4P2.
First of all it is not the question of combination, it is of permutation (assumed : letters without repetition).
Reason : suppose there are two letters I want to arrange le t say they are t and p, so tp and pt conveys the different words.
So order matters here, so we use 7P3*4P2 instead of 7C3*4C2.
So the answer is 5!*7P3*4P2.
K ANIRUDH said:
8 years ago
Hi, @All.
I am just confused with committee problems and this question in committee problems we are not doing this 5! thing if a committee of 5 people is to be made, we are just selecting the men and women out of given men and women and multiplying why it is not similar in this case.
Can anyone help to understand the problem? Please help me.
I am just confused with committee problems and this question in committee problems we are not doing this 5! thing if a committee of 5 people is to be made, we are just selecting the men and women out of given men and women and multiplying why it is not similar in this case.
Can anyone help to understand the problem? Please help me.
(5)
Nitish said:
9 years ago
As the consonants and vowels are not specified and we only have to find how many ways we can arrange them i.e. 210 the Order does not matter. If the letters were specified then only we can apply permutation here.
For example, if the question was 7 men and 4 women, we wouldn't multiply with 5!.
Hence, the correct answer would be 210.
For example, if the question was 7 men and 4 women, we wouldn't multiply with 5!.
Hence, the correct answer would be 210.
Liam Martin said:
6 years ago
Here we can directly permute 7p3*4p2=2520 so this gives us 3 vowels say "aei" and 2 consonants say bc so aeibc we can then calculate the number of groups made by the 2 consonants or 3 values from this group of 5 letters chosen in total so the solution is 7p3*4p2*5c2 = 7*6*5*4*3*(5*4)/2=2520 *10 = 25200.
(4)
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.
Burt Yellin said:
8 years ago
I'm not sure this is the correct forum for this question, but here goes. If I have 4 consonants and there are 5 vowel sounds that can be pronounced hard or soft, how many different pronunciations are possible?
If this is the wrong forum can you direct me to a place that could answer that question?
If this is the wrong forum can you direct me to a place that could answer that question?
Ridsie said:
1 decade ago
A bag contains a total at 105 coins of Re.1, 50 paise and 25 paise denominations. Find the total number of coins of Re.1, if there are total number of 50.50 rupees in the bag and it is known that the number of 25 paise coins is 133.33% more than Re.1 coins.
Mark Burnham said:
8 years ago
This is a question of permutation, as the order in which letters are arranged changes the final result.
Therefore = 7P3*4P2 = 2520.
However, there are 10 different ways that the consonants and vowels can be arranged.
So, the answer is 2520*10,
=2520.
Therefore = 7P3*4P2 = 2520.
However, there are 10 different ways that the consonants and vowels can be arranged.
So, the answer is 2520*10,
=2520.
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