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?
210
1050
25200
21400
None of these
Answer: Option
Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

      = (7C3 x 4C2)
= 7 x 6 x 5 x 4 x 3
3 x 2 x 1 2 x 1
= 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 4 of 7.

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.

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.

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.

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!

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.

Prof. Vedavyas said:   9 years ago
Explanation is actually simple. Suppose if they asked to find the number of words possible using five different letters and you had no constraints of consonants and vowels.

Then the answer would obviously 5!. So, we know you can form 5! words from five letters.

Now, in the problem, there are not just 5 letters, but much more - 7 consonants and 4 vowels. The 3 consonants can be selected in 7C3 ways. For each of these, we can select 2 vowels in 4C2 ways.

So, the 5 letters themselves can be selected in 7C3X4C2 ways.

Note that each of these 7C3X4C2 are a different set of five letters. A given set of 5 letters can be arranged in 5! ways.

So, 7C3X4C2 sets of 5 letters can be arranged in 7C3X4C2X5! ways.

Hence the answer is 25,200


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