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 7 of 7.
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.
Athar said:
1 decade ago
You have to arrange the letter.
3 consonants out of 7.
2 vowels out of 4.
7c3x4c2 = 210.
You can do,
How many letter out of these letters?
3+2 = 5.
Now the required no of ways = 210 x 5!
3 consonants out of 7.
2 vowels out of 4.
7c3x4c2 = 210.
You can do,
How many letter out of these letters?
3+2 = 5.
Now the required no of ways = 210 x 5!
Rob_ph said:
1 decade ago
210 is correct. What is asked in the problem is how many words that have 3 consonants and 2 vowels can be formed from the given 7 and 4.
Basically all words that contains 3 consonants no matter how you arranged the 3 consonants that you picked is considered as one, same with the vowels. Meaning QEWAR is same with WEQAR, hence counted as 1 sample.
Unless you choose another consonant say QEWAD or another vowel QIWAR then it will be a different sample.
Basically all words that contains 3 consonants no matter how you arranged the 3 consonants that you picked is considered as one, same with the vowels. Meaning QEWAR is same with WEQAR, hence counted as 1 sample.
Unless you choose another consonant say QEWAD or another vowel QIWAR then it will be a different sample.
Leonardo said:
10 years ago
I don't understand it.
Rishi said:
10 years ago
In question instead of how many words there should be how many arrangements.
Otherwise, a solution is wrong according to question.
Otherwise, a solution is wrong according to question.
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