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# S99-14

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Math Expert
Joined: 02 Sep 2009
Posts: 57025

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16 Sep 2014, 01:53
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Difficulty:

95% (hard)

Question Stats:

37% (01:40) correct 63% (01:41) wrong based on 89 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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Math Expert
Joined: 02 Sep 2009
Posts: 57025

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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 $$3 \times 2 \times 3 \times 2 \times 3 = 108$$.

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Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 410
Location: Netherlands
GMAT Date: 02-11-2015
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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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Joined: 25 Nov 2014
Posts: 155
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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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Joined: 12 Oct 2015
Posts: 5
GMAT 1: 740 Q50 V40

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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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Location: India
Concentration: Operations, Entrepreneurship
GMAT 1: 610 Q49 V25
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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..
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+1 Kudo, if my post helped
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29 Aug 2016, 03:56
The word value tricked me. I guess it has to be "vowel"
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Joined: 24 Feb 2017
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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
Math Expert
Joined: 02 Sep 2009
Posts: 57025

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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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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.

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08 Jan 2018, 16:31
'values" should be "vowels"
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Kindly press the +1Kudos if you like the explanation. Thanks a lot!!!
Intern
Joined: 27 Jul 2019
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03 Aug 2019, 19:22
Bunuel wrote:
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 $$3 \times 2 \times 3 \times 2 \times 3 = 108$$.

Hi Bunuel,

I wanted to clarify the combinatorics method that you used, so when it is CVCVC and 3 possibilities for C and 2 for V, did you do 3C1x2C1x3C1x2C1x3C1 = 108?

Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 57025

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03 Aug 2019, 23:11
Tlo19 wrote:
Bunuel wrote:
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 $$3 \times 2 \times 3 \times 2 \times 3 = 108$$.

Hi Bunuel,

I wanted to clarify the combinatorics method that you used, so when it is CVCVC and 3 possibilities for C and 2 for V, did you do 3C1x2C1x3C1x2C1x3C1 = 108?

Thanks!

That's not my solution. It's MGMAT's. My solution is here: https://gmatclub.com/forum/s99-184742.html#p1885348
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Re: S99-14   [#permalink] 03 Aug 2019, 23:11
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# S99-14

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