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The Simplastic language has only 2 unique values and 3

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

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New post 16 Aug 2017, 22:15
shashankism wrote:
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...


Thanks for your reply Shashankism, Is it not 120?, below is my reasoning Please correct me where I am wrong

IT is CVCVC
Considering Consonants
3 Consonants 1st C in 3 ways, 2nd C in 2 Ways and Third C in 1 Ways = 3*2*1 = 6 ways

Considering Vowels
Out of 5 vowels 2 vowels can be selected in 5C2 = 10 ways and it(2 vowels) can be permuted 2! ways which 20

Hence it is 6*20 = 120 ways
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Re: The Simplastic language has only 2 unique values and 3  [#permalink]

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New post 16 Aug 2017, 22:57
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joepc wrote:
shashankism wrote:
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...


Thanks for your reply Shashankism, Is it not 120?, below is my reasoning Please correct me where I am wrong

IT is CVCVC
Considering Consonants
3 Consonants 1st C in 3 ways, 2nd C in 2 Ways and Third C in 1 Ways = 3*2*1 = 6 ways

Considering Vowels
Out of 5 vowels 2 vowels can be selected in 5C2 = 10 ways and it(2 vowels) can be permuted 2! ways which 20

Hence it is 6*20 = 120 ways


But the question stem says there are only 2 vowels(check question it says there are 2 unique vowels misspelled as values and 3 unique consonant)..
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Re: The Simplastic language has only 2 unique values and 3  [#permalink]

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New post 12 Feb 2018, 02:58
VeritasPrepKarishma wrote:
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 ... ts-part-i/
http://www.veritasprep.com/blog/2011/10 ... s-part-ii/


Hi Karishma

I just read through your post and solved your Questions in the blog post. However I am a bit confused now, let's take Q2 from your blogpost:

Your solution: 10*9*8*7*6= 30'240
My approach: 10!/5!*(10-5)!=252

What is the difference in our two approaches? When are we supposed to use which one?

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

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New post 12 Feb 2018, 08:05
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


Nothing mentioned about repetition. so repetition allowed.

3*2*3*2*3= 108
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Re: The Simplastic language has only 2 unique values and 3  [#permalink]

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New post 12 Feb 2018, 21:32
Gregsterh wrote:
VeritasPrepKarishma wrote:
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 ... ts-part-i/
http://www.veritasprep.com/blog/2011/10 ... s-part-ii/


Hi Karishma

I just read through your post and solved your Questions in the blog post. However I am a bit confused now, let's take Q2 from your blogpost:

Your solution: 10*9*8*7*6= 30'240
My approach: 10!/5!*(10-5)!=252

What is the difference in our two approaches? When are we supposed to use which one?

Thanks


A password is a permutation (or arrangement) problem. So ABDFG is different from BDAFG.
You have used 10C5 which is the number of ways in which we SELECT the 5 letters to use. We also need to ARRANGE them so we need to multiply by 5! to get
[10!/5!*(10-5)!] * 5! = 10!/5! = 10*9*8*7*6
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The Simplastic language has only 2 unique values and 3  [#permalink]

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New post 12 Feb 2018, 21:52
In Question there is not mentioned that repetition is not allowed , so we can assume that same vowel or consonant can occur multiple times but with different noun .
That's why solution must be like that -
C V C V C
3*2*3*2*3 =108.
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Re: The Simplastic language has only 2 unique values and 3  [#permalink]

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New post 05 Mar 2018, 11:42
Hi All,

The question you posted has a small typo in it. I believe it's means to say….

2 unique Vowels
3 unique Consonants

We're asked for all of the various 5 letter "nouns" that follow the pattern CVCVC (in which C is a consonant and V is a vowel) and that could occur in this language.

This is essentially just a permutation question. Since there are 3 different consonants and 2 different vowels, we would end up with…

(3)(2)(3)(2)(3) = 108 different 5 letter "nouns"

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

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New post 20 Nov 2018, 18:54
CVCVC
There are 108 different ways of selecting CVCVC (in this format only).
Event of selecting first consonant is independent of selecting Vowvel for 2nd event.
Thats's why we are multiplying 3*2*3*2*3
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Re: The Simplastic language has only 2 unique values and 3 &nbs [#permalink] 20 Nov 2018, 18:54

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