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Combinatorics

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Combinatorics

by dev.gavande » Wed Oct 05, 2011 2:00 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?
"¢ 9
"¢ 12
"¢ 36
"¢ 72
"¢ 108

I was unable to solve, cannot understand the explanation as well.
Source : MGMAT
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by shankar.ashwin » Wed Oct 05, 2011 2:07 am
FIx the positions of 2 vowels, the 3 consonants can be arranged in 3! ways = 6

And the vowels can be interchanged in the 2 positions, so 6*2 =12
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by sl750 » Wed Oct 05, 2011 2:08 am
The first consonant can be any of the 3 consonants according to the principles of this language, similarly the second consonant can be any of the remaining 2 consonants and so on
The first vowel can be any of the 2 available vowels
So we have 3*2*2*1*1 = 12 unique nouns
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by shankar.ashwin » Wed Oct 05, 2011 2:13 am
I see repetition can be allowed here,

So you have 3*2*3*2*3 = 108 (3 possibilities for consonants and 2 for vowels)
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