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Simplastic

by j_shreyans » Fri Sep 19, 2014 9:04 am
The Simplastic language has only 2 unique values 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?

A)9
B)12
C)36
D)72
E)108

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by [email protected] » Fri Sep 19, 2014 10:53 am
Hi j_shreyans,

The question you posted has a small typo in it. I believe it's means to say....

2 unique Vowels
3 unique Consonants

We're asked for all of the various 5 letter "nouns" that follow the pattern CVCVC (in which C is a consonant and V is a vowel) and that could occur in this language.

This is essentially just a permutation question. Since there are 3 different consonants and 2 different vowels, we would end up with...

(3)(2)(3)(2)(3) = 108 different 5 letter "nouns"

Final Answer: E

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by Brent@GMATPrepNow » Fri Sep 19, 2014 12:50 pm
j_shreyans wrote: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?

A)9
B)12
C)36
D)72
E)108

OAE
NOTE: based on the OA, I'm assuming that it's okay to use a letter more than once.

Take the task of creating nouns and break it into stages.

Stage 1: Choose a consonant for the 1st letter.
There are 3 consonants to choose from, so, we can complete stage 1 in 3 ways

Stage 2: Choose a vowel for the 2nd letter.
There are 2 vowels to choose from, so, we can complete stage 2 in 2 ways

Stage 3: Choose a consonant for the 3rd letter.
There are 3 consonants to choose from, so, we can complete stage 3 in 3 ways

Stage 4: Choose a vowel for the 4th letter.
There are 2 vowels to choose from, so, we can complete stage 4 in 2 ways

Stage 5: Choose a consonant for the 5th letter.
There are 3 consonants to choose from, so, we can complete stage 5 in 3 ways

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus create a noun) in (3)(2)(3)(2)(3) ways ([spoiler]= 108 ways[/spoiler])

Answer: E
--------------------------

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Cheers,
Brent
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by RiyaR » Sat Sep 27, 2014 4:49 am
If no repetition were to be allowed, would the number of eays be 12?
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by [email protected] » Sat Sep 27, 2014 1:12 pm
Hi RiraR,

Yes, if duplicate letters were NOT allowed, the math would be:

(3)(2)(2)(1)(1) = 12 different "words"

You'll notice that the math "steps" are remarkably similar to what we did to answer the original question. These types of questions are ultimately about following the "rules" that the question lays out.

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