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S99-14

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S99-14 [#permalink]

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New post 16 Sep 2014, 01:53
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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

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Re S99-14 [#permalink]

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New post 16 Sep 2014, 01:53
Official Solution:

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


This Combinatorics problem asks you to compute the number of possible nouns in Simplastic, given the template CVCVC.

We have a series of successive choices:
  • Pick the first consonant
  • Pick the first vowel
  • Pick the second consonant
  • Pick the second vowel
  • Pick the third consonant

So we need to count the choices we have at each stage, and then multiply these choices together. We have 3 choices for each consonant and 2 choices for each vowel. Note that we can reuse consonants and vowels. For instance, imagine that the consonants are {g, l, t} and the vowels are {a, u}. Here are some valid nouns in Simplastic:

gagag

gulat

lugul

Thus, we write \(3 \times 2 \times 3 \times 2 \times 3 = 108\).


Answer: E
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Re: S99-14 [#permalink]

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New post 07 Jul 2015, 11:23
Same way, using the slot method:

We need the nouns to be of the form CVCVC

We draw the slots:
__....__....__...__...__

Now, we have 3 options for C and 2 options for V and the slots will be filled in like this:
3....2....3....2....3 --> 3^3*2^2 = 108. ANS E

This is easily seen if you pick letters for the Cs and the Vs:
Cs will be B,C,D
Vs will be A,E

The slots now become:
C....V....C....V....C
B...A....B....A....B
C...E....C....E....C
D........D...........D

However, these are only some of the options, as the Cs and the Vs can be swapped. So, instead of B being first, C r D can be first, and instead of A being first E can be first. In other words, there are 3*2 options for the first pair of C and V and 3*2 options for the second pair of C and V and 3 remaining options for the last C, leading to 6*6*3 = 108.
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Re: S99-14 [#permalink]

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New post 30 Jul 2015, 20:44
good ques.... tricked me into selecting 12. I didnt notice that repetition is allowed. Silly mistake.
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Re S99-14 [#permalink]

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New post 12 Apr 2016, 12:29
I think this is a high-quality question and I agree with explanation. Hello!

I think the first stated "value" should be a "vowel"
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Re: S99-14 [#permalink]

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New post 26 Apr 2016, 09:05
AA2014 wrote:
good ques.... tricked me into selecting 12. I didnt notice that repetition is allowed. Silly mistake.

Same mistake mate.. :cry:
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Re: S99-14 [#permalink]

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New post 29 Aug 2016, 03:56
The word value tricked me. I guess it has to be "vowel"
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S99-14 [#permalink]

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New post 10 Jul 2017, 03:04
I'm wondering why the initial selection of the consonants and vowels didn't factor into the calculation. I agree that such a high number isn't in any of the options.

The initial 3 consonants and 2 vowels can be any of the 21 and 5 consonants and vowels resp. in the english alphabet? I think the only thing that makes this wrong is the assumption that the Vowels and Consonants in Simplastic is taken from English, and it doesn't say that anywhere.

I got 21C3 X 5C2 X 108
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Re: S99-14 [#permalink]

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New post 10 Jul 2017, 03:23
rbramkumar wrote:
I'm wondering why the initial selection of the consonants and vowels didn't factor into the calculation. I agree that such a high number isn't in any of the options.

The initial 3 consonants and 2 vowels can be any of the 21 and 5 consonants and vowels resp. in the english alphabet? I think the only thing that makes this wrong is the assumption that the Vowels and Consonants in Simplastic is taken from English, and it doesn't say that anywhere.

I got 21C3 X 5C2 X 108


Because it's Simplastic language (which has 2 unique vowels and 3 unique consonants only), not English.
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Re: S99-14 [#permalink]

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New post 10 Jul 2017, 03:24
Bunuel 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


Alternation solution:

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.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: S99-14 [#permalink]

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New post 08 Jan 2018, 16:31
'values" should be "vowels"
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Re: S99-14   [#permalink] 08 Jan 2018, 16:31
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