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They did not say that C and V cannot be repeated hence it is 3*2*3*2*3 = 27*4=108

Answer E
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KarishmaB

I have the same question. Would you be able to help me out here please.
When do we divide by the number of repetitions in an arrangement, and why are we not doing that in this case?

Your responses on this post have been extremely helpful and easy to internalize.

Thanks in advance!

honchos



Great questions, at one point when this situation will arise, wouldn't we divide it by -

3*2*3*2*3/3!X2!
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Had we been calculating the number of arrangements of 3Cs and 2Vs, then we would divided by 3!*2!. But we are not calculating CCCVV, CVCCV etc.
We are only calculating the number of different ways in which we can write a Consonant, then a Vowel, then consonant, then vowel and then consonant again.
Say consonants are D, F, G and vowels are A, E (for convenience), then we are calculating DADAD, DAFAG, FEGAD etc. Each time it is CVCVC only.

GyanviP
KarishmaB

I have the same question. Would you be able to help me out here please.
When do we divide by the number of repetitions in an arrangement, and why are we not doing that in this case?

Your responses on this post have been extremely helpful and easy to internalize.

Thanks in advance!


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@bunuel - IN this question , I was assuming there were 21 consonants and 5 vovwels and then solving accordingly. Why did you choose 3 and 2 for consonants and vowels respectively and why not 21 and 5 ? Since there are 21 consonants and 5 vowels
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SaumyaKhare
@bunuel - IN this question , I was assuming there were 21 consonants and 5 vovwels and then solving accordingly. Why did you choose 3 and 2 for consonants and vowels respectively and why not 21 and 5 ? Since there are 21 consonants and 5 vowels
The Simplastic language has only 2 unique vowels and 3 unique consonants. Every noun in Simplastic has the structure CVCVC, where C stands for a consonant and V stands for a vowel. How many different nouns are possible in Simplastic?
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KarishmaB so whenever we get questions like different ways than like an arrangement, we don't have to do the repetition division? can i take it like that?
KarishmaB
Had we been calculating the number of arrangements of 3Cs and 2Vs, then we would divided by 3!*2!. But we are not calculating CCCVV, CVCCV etc.
We are only calculating the number of different ways in which we can write a Consonant, then a Vowel, then consonant, then vowel and then consonant again.
Say consonants are D, F, G and vowels are A, E (for convenience), then we are calculating DADAD, DAFAG, FEGAD etc. Each time it is CVCVC only.


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One word in the question changes it completely. Such rules don't work. You need to see exactly what is asked.

SwethaReddyL
KarishmaB so whenever we get questions like different ways than like an arrangement, we don't have to do the repetition division? can i take it like that?

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