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# The Simplastic language has only 2 unique values and 3

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The Simplastic language has only 2 unique values and 3 [#permalink]  06 Dec 2010, 03:17
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jun 2012, 03:28, edited 1 time in total.
Edited the question
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Re: MGMAT Challenge Test 1 #14 [#permalink]  06 Dec 2010, 03:19
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The answer is E because order does not matter making the combination 3*2*3*2*3?
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Re: MGMAT Challenge Test 1 #14 [#permalink]  06 Dec 2010, 03:44
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
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Re: MGMAT Challenge Test 1 #14 [#permalink]  06 Dec 2010, 04:46
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mmcooley33 wrote:
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

mmcooley33 wrote:
The answer is E because order does not matter making the combination 3*2*3*2*3?

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.

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.

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

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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Re: MGMAT Challenge Test 1 #14 [#permalink]  06 Dec 2010, 05:59
bunuel, what concept is this testing? I seem to not get the 3*2*3*2*3 aspect of your solution.
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Re: MGMAT Challenge Test 1 #14 [#permalink]  06 Dec 2010, 06:11
Expert's post
anish319 wrote:
bunuel, what concept is this testing? I seem to not get the 3*2*3*2*3 aspect of your solution.

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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Re: MGMAT Challenge Test 1 #14 [#permalink]  18 Sep 2013, 03:18
Bunuel wrote:
mmcooley33 wrote:
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

mmcooley33 wrote:
The answer is E because order does not matter making the combination 3*2*3*2*3?

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.

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.

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

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.

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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The Simplastic language [#permalink]  20 Oct 2013, 05:03
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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Re: The Simplastic language [#permalink]  20 Oct 2013, 05:04
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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Re: The Simplastic language [#permalink]  20 Oct 2013, 05:08
Expert's post
SaraLotfy wrote:
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

Merging similar topics. Please refer to the solutions above.
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Re: The Simplastic language [#permalink]  20 Oct 2013, 05:09
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Re: The Simplastic language has only 2 unique values and 3 [#permalink]  04 Oct 2014, 12:14
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.
Re: The Simplastic language has only 2 unique values and 3   [#permalink] 04 Oct 2014, 12:14
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