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.
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?
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.
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.
David said:
1 decade ago
Simplify problem.
3 consonants BCD.
3 vowels AEI.
Choose 2 consonants, no repetition.
Choose 2 vowels, no repetition.
Form different words from the chosen consonants and vowels.
Choose 2 consonants, 3 ways - BC, BD, CD.
Choose 2 vowels, 3 ways - AE, AI, EI.
Different groups with 2 consonants and 2 vowels, 3x3 = 9.
BCAE, BCAI, BCEI, BDAI, BDAI, BDEI, CDAE, CDAI, CDEI.
Different words from 4 character set such as BCAE, 4! = 4x3x2x1 = 24.
BCAE, BCEA, BACE, BAEC, BECA, BEAC.
CAEB, CABE, CEAB, CEBA, CBAE, CBEA.
ABCE, ABEC, AEBC, AECB, ACBE, ACEB.
EBCA, EBAC, ECBA, ECAB, EABC, EACB.
Total 4 character words from 2 different consonants and 2 different vowels would be 9x24 = 216.
Apply to above.
7C3 x 4C2 x 5! = 25200.
More difficult would be when repetitions are allowed in consonants and vowels.
3 consonants BCD.
3 vowels AEI.
Choose 2 consonants, no repetition.
Choose 2 vowels, no repetition.
Form different words from the chosen consonants and vowels.
Choose 2 consonants, 3 ways - BC, BD, CD.
Choose 2 vowels, 3 ways - AE, AI, EI.
Different groups with 2 consonants and 2 vowels, 3x3 = 9.
BCAE, BCAI, BCEI, BDAI, BDAI, BDEI, CDAE, CDAI, CDEI.
Different words from 4 character set such as BCAE, 4! = 4x3x2x1 = 24.
BCAE, BCEA, BACE, BAEC, BECA, BEAC.
CAEB, CABE, CEAB, CEBA, CBAE, CBEA.
ABCE, ABEC, AEBC, AECB, ACBE, ACEB.
EBCA, EBAC, ECBA, ECAB, EABC, EACB.
Total 4 character words from 2 different consonants and 2 different vowels would be 9x24 = 216.
Apply to above.
7C3 x 4C2 x 5! = 25200.
More difficult would be when repetitions are allowed in consonants and vowels.
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