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The Simplastic language has only 2 unique values and 3 [#permalink]
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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
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Last edited by Bunuel on 17 Jun 2012, 03:28, edited 1 time in total.
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Re: MGMAT Challenge Test 1 #14 [#permalink]
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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]
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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]
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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 CVCVC. 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. It's basically the same if it were how many different 5digit numbers are possible with the following structure oddevenoddevenodd, where odd numbers can be only 1, 3 or 5 and even numbers only 2 and 4. adhithya wrote: 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 (XAYAZ) or 3 same consonants (XAXAX), so if this was not the case then this would be explicitly mentioned.
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Re: MGMAT Challenge Test 1 #14 [#permalink]
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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]
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06 Dec 2010, 06:11
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, 2letter code ConsonantVowel, 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]
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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 CVCVC. 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. It's basically the same if it were how many different 5digit numbers are possible with the following structure oddevenoddevenodd, where odd numbers can be only 1, 3 or 5 and even numbers only 2 and 4. adhithya wrote: 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 (XAYAZ) or 3 same consonants (XAXAX), 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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Re: The Simplastic language [#permalink]
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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]
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20 Oct 2013, 05:09



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Re: The Simplastic language has only 2 unique values and 3 [#permalink]
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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 reused.



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Re: The Simplastic language has only 2 unique values and 3 [#permalink]
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26 May 2015, 18:59
Hi, do we need to account for the restriction that impose by the structure as CVCVC? as one V need to follow a C and we cant do CCCVV? I am confused here..
And on top of it, when is it good to use the formula of combination and when we just use the method applied in this question ( like a number lock), thought believe that its the same concept?
Thank you..



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Re: The Simplastic language has only 2 unique values and 3 [#permalink]
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26 May 2015, 19:35
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katzzzz wrote: Hi, do we need to account for the restriction that impose by the structure as CVCVC? as one V need to follow a C and we cant do CCCVV? I am confused here..
And on top of it, when is it good to use the formula of combination and when we just use the method applied in this question ( like a number lock), thought believe that its the same concept?
Thank you.. We are accounting for it by calculating only the number of ways of writing CVCVC. So the other arrangements of 3Cs and 2Vs are ignored. You can write the first C in 3 ways. You can write the next letter V in 2 ways. The next letter is again C for which we again have 3 options (note that repetition of letters is not a problem) The next letter V can be chosen in 2 ways. The last letter C can be chosen in 3 ways again. This gives us 3*2*3*2*3 = 108 ways. You use the combination formula only when you have to select a few things out of many things. Here, no selection is required. Say, if there were 10 consonants and we had to make the nouns using 3 DISTINCT consonants, then we would have SELECTED 3 of the 10 (in 10C3 ways) and then arranged them in 3 places in 3! ways. The method used in this question is the basic counting principle. It is used when you have distinct places for things. I suggest you to check out these posts: http://www.veritasprep.com/blog/2011/10 ... inatorics/http://www.veritasprep.com/blog/2011/11 ... binations/
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Re: The Simplastic language has only 2 unique values and 3 [#permalink]
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27 May 2015, 04:07
thank you! it really helps! for special seating arrangement ( A must proceed by B), guess we use the same approach here rather than the formula as we don't have to choose something from a group?



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Re: The Simplastic language has only 2 unique values and 3 [#permalink]
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katzzzz wrote: thank you! it really helps! for special seating arrangement ( A must proceed by B), guess we use the same approach here rather than the formula as we don't have to choose something from a group? Yes, you use this concept for arrangements. Here is how you solve linear arrangements with constraints: http://www.veritasprep.com/blog/2011/10 ... tsparti/http://www.veritasprep.com/blog/2011/10 ... spartii/
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Re: The Simplastic language has only 2 unique values and 3 [#permalink]
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15 Jul 2015, 05:21
Hi Bunuel , for any question on combination in GMAT . if no condition is stated then can we say that ... repetition is allowed . Please confirm .
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Re: The Simplastic language has only 2 unique values and 3 [#permalink]
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15 Jul 2015, 07:33
Vowels  2, Consonants  3. cvcvc = 3*2*3*2*3 = 108. Ans (E).
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Re: The Simplastic language has only 2 unique values and 3 [#permalink]
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16 Aug 2017, 11:08
TheKingInTheNorth wrote: Hi Bunuel ,
for any question on combination in GMAT . if no condition is stated then can we say that ... repetition is allowed .
Please confirm . Yes if the question is not providing any restriction on repetition means repetition is allowed... Also this is logical as we have repetition allowed in forming word (for e.g. rEpETITIon.)
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Re: The Simplastic language has only 2 unique values and 3 [#permalink]
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16 Aug 2017, 11:11
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 Since the noun has a structure CVCVC 1st alphabet C can take 3 values 2nd alphabet V can take 2 values 3rd alphabet C can take 3 values 4th alphabet V can take 2 values 5th alphabet C can take 3 values So, different nouns possible = 3*2*3*2*3 = 108 Answer E
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Re: The Simplastic language has only 2 unique values and 3 [#permalink]
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16 Aug 2017, 20:31
Dear Math Experts,
What will be answer if repetition in CVCVC is not allowed(once a vowel or consonant is used it cannot be used again)
Thanks Joepc



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Re: The Simplastic language has only 2 unique values and 3 [#permalink]
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16 Aug 2017, 21:18
joepc wrote: Dear Math Experts,
What will be answer if repetition in CVCVC is not allowed(once a vowel or consonant is used it cannot be used again)
Thanks Joepc In that case 1st C can take 3 values, 2nd C can take 2 values and 3rd C can take 1 value. Similarly 1st V can take 2 values and 2nd V van take 1 value.. So, total no. of words formed = 3*2*1*2*1 = 12...
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Re: The Simplastic language has only 2 unique values and 3
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