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 6 of 7.
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.
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.
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.
Samiksha said:
1 decade ago
@Ridsie.
Let the no. of Re. 1 coin be x.
==>no. of 25 paise coins = x+1.33 x = 2.33 x.
==>no. of 50 paise coins = 105-(x+2.33 x) = 105-3.33 x.
(1)x+(25*2.33 x)/60+(50*(105-3.33 x))/60 = 50.50.
Solve for x.
Let the no. of Re. 1 coin be x.
==>no. of 25 paise coins = x+1.33 x = 2.33 x.
==>no. of 50 paise coins = 105-(x+2.33 x) = 105-3.33 x.
(1)x+(25*2.33 x)/60+(50*(105-3.33 x))/60 = 50.50.
Solve for x.
Spurthy said:
1 decade ago
I cannot get you, after what you said after 120 ways?
Neha said:
1 decade ago
Why we multiply with 5! I cant understand. If we multiply with 5! in words then why we not multiply in persons it will also formed in different manner. Please someone explain this.
Ajay said:
1 decade ago
Can anyone help me to understand when have to take NCR or when have to take NPR in easily manner. I am more confused about this concept.
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