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# S99-14

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:53
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Difficulty:

95% (hard)

Question Stats:

36% (01:29) correct 64% (01:17) wrong based on 116 sessions

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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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16 Sep 2014, 00:53
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 $$3 \times 2 \times 3 \times 2 \times 3 = 108$$.

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07 Jul 2015, 10:23
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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30 Jul 2015, 19:44
good ques.... tricked me into selecting 12. I didnt notice that repetition is allowed. Silly mistake.
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Joined: 12 Oct 2015
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12 Apr 2016, 11:29
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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26 Apr 2016, 08:05
AA2014 wrote:
good ques.... tricked me into selecting 12. I didnt notice that repetition is allowed. Silly mistake.

Same mistake mate..
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+1 Kudo, if my post helped

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29 Aug 2016, 02:56
The word value tricked me. I guess it has to be "vowel"
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10 Jul 2017, 02:04
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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10 Jul 2017, 02:23
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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10 Jul 2017, 02:24
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.

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08 Jan 2018, 15:31
'values" should be "vowels"
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Kindly press the +1Kudos if you like the explanation. Thanks a lot!!!

Re: S99-14 &nbs [#permalink] 08 Jan 2018, 15:31
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# S99-14

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