The nouns have fixed structure C-V-C-V-C. Now, each C can take 3 values (let's say X, Y or Z) and each V can take 2 values (let's say A or E), so there will be 3*2*3*2*3=108 nouns possible.

Answer: E.

It's basically the same if it were how many different 5-digit numbers are possible with the following structure odd-even-odd-even-odd, where odd numbers can be only 1, 3 or 5 and even numbers only 2 and 4.



It's natural to think that a noun can have for example two same vowels (X-A-Y-A-Z) or 3 same consonants (X-A-X-A-X), so if this was not the case then this would be explicitly mentioned.
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108 is the answer if we assume that repetition is allowed,but how do we know whether repn is allowed or not if question doesnt mention anything

Consider simpler case, 2-letter code Consonant-Vowel, where we can use only B, C or D for a consonant (3 options) and only A or E for a vowel (2 options). How many codes are possible?

BA;
BE;
CA;
CE;
DA;
DE.

So, total of 6 codes, 3*2=6, are possible. This is called Principle of Multiplication: If one event can occur in \(m\) ways and a second can occur independently of the first in \(n\) ways, then the two events can occur in \(mn\) ways.

Now, the above is just expanded to CVCVC structure in the original question.

Hope it's clear.
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My approach was as follows: 2 values and 3 constants can be counted as 2! * 3! = 12. I treated it as a counting problem with repeated values, why is this wrong?
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Answer is E. It just says any combination of CVCVC will work and so, calculate 3*2*3*2*3=108.

Doesn't mean 3*2*2*1*1, where you should subtract 1 each time. The letters can be re-used.

We are accounting for it by calculating only the number of ways of writing CVCVC. So the other arrangements of 3Cs and 2Vs are ignored.
You can write the first C in 3 ways.
You can write the next letter V in 2 ways.
The next letter is again C for which we again have 3 options (note that repetition of letters is not a problem)
The next letter V can be chosen in 2 ways.
The last letter C can be chosen in 3 ways again.
This gives us 3*2*3*2*3 = 108 ways.

You use the combination formula only when you have to select a few things out of many things. Here, no selection is required. Say, if there were 10 consonants and we had to make the nouns using 3 DISTINCT consonants, then we would have SELECTED 3 of the 10 (in 10C3 ways) and then arranged them in 3 places in 3! ways.
The method used in this question is the basic counting principle. It is used when you have distinct places for things. I suggest you to check out these posts:
http://www.veritasprep.com/blog/2011/10 ... inatorics/
http://www.veritasprep.com/blog/2011/11 ... binations/
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Yes, you use this concept for arrangements. Here is how you solve linear arrangements with constraints:

http://www.veritasprep.com/blog/2011/10 ... ts-part-i/
http://www.veritasprep.com/blog/2011/10 ... s-part-ii/
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Vowels - 2, Consonants - 3.

c-v-c-v-c = 3*2*3*2*3 = 108. Ans (E).
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Since the noun has a structure CVCVC
1st alphabet C can take 3 values
2nd alphabet V can take 2 values
3rd alphabet C can take 3 values
4th alphabet V can take 2 values
5th alphabet C can take 3 values

So, different nouns possible = 3*2*3*2*3 = 108

Answer E
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In that case 1st C can take 3 values, 2nd C can take 2 values and 3rd C can take 1 value.

Similarly 1st V can take 2 values and 2nd V van take 1 value..

So, total no. of words formed = 3*2*1*2*1 = 12...
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