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.
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.
Jyoti bala said:
1 decade ago
When we apply permutation and combination? please tell me one.
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
Jatin Bavishi said:
1 decade ago
This cannot be a permutation problem because the first thing we ought to do is select 4 consonants out of 7 and 2 vowels out of 4.
Let us approach the question by assigning actual alphabets, and reducing the quantity. Suppose there are 3 consonants B, C and D and 2 vowels A and E, and we have to form a word using 2 consonants and 1 vowel.
The required words are: ABC, BAC, BCA, ACB, CAB, CBA, ACD, CAD, CDA, DCA, DAC, ADC.....the sum would be 36.
Let us now do the sum using formula:
3P2*2P1=12, which is clearly not the answer.
Let us approach the question by assigning actual alphabets, and reducing the quantity. Suppose there are 3 consonants B, C and D and 2 vowels A and E, and we have to form a word using 2 consonants and 1 vowel.
The required words are: ABC, BAC, BCA, ACB, CAB, CBA, ACD, CAD, CDA, DCA, DAC, ADC.....the sum would be 36.
Let us now do the sum using formula:
3P2*2P1=12, which is clearly not the answer.
Sagar said:
1 decade ago
If it was mentioned distinct in the question can we use 7p3*4p2.
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.
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?
Mansi said:
1 decade ago
What is the difference between permutation and combination? I always get confused.
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.
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