The Simplastic language has only 2 unique vowels and 3 unique consonan

The answer is E because order does not matter making the combination 3*2*3*2*3?

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

bunuel, what concept is this testing? I seem to not get the 3*2*3*2*3 aspect of your solution.

Consider simpler case, 2-letter code Consonant-Vowel, 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.

Great questions, at one point when this situation will arise, wouldn't we divide it by -

3*2*3*2*3/3!X2!

Hi, do we need to account for the restriction that impose by the structure as C-V-C-V-C? as one V need to follow a C and we cant do C-C-C-V-V?
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 videos:
Video on Permutations: https://youtu.be/LFnLKx06EMU
Video on Combinations: https://youtu.be/tUPJhcUxllQ

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:

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/10 ... ts-part-i/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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

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"

Final Answer:

GMAT assassins aren't born, they're made,
Rich

The number of nouns are possible in Simplastic is 3 x 2 x 3 x 2 x 3 = 27 x 4 = 108.

Answer: E

Bunuel

Please explain why the permutation formula would not work here. What cases would it be missing?

I understand how the idea of 3*2*3*2*3 works, but I used the permutation formula and got the wrong answer:

3P3 * 2P2 = 3! * 2! = 12

Hi testtakerstrategy,

The prompt does NOT state that we can only use each letter once. For example, there's a difference between all of the ways to arrange 3 different letters (that would be 3! = 6) and all of the three-letter 'words' you can form with 3 different letters in which you can use a letter more than once (that would be (3)(3)(3) = 27).

GMAT assassins aren't born, they're made,
Rich

The quesion is asking total number of nouns ...

So out of 21 consonant we can pick any 3 and out of 5 we can pick any 2 ....

going by this logic ... why can't the answer be 21P3 * 5P2 ???

VeritasKarishma / Bunuel /- experts please help!!

Hi razorhell,

The prompt provides some specific 'restrictions' that we have to follow:
1) The 'language' has just 2 vowels and just 3 consonants.
2) Every noun in this language is written in the same 'structure': CVCVC (where C is a consonant and V is a vowel).
3) There are NO restrictions as to how many times you can use a particular letter (re: the 2 vowels in a noun could be the SAME vowel).

Since we're putting letters in a specific order to form nouns, that is a PERMUTATION. The total number of nouns that fit these rules is (3)(2)(3)(2)(3).

GMAT assassins aren't born, they're made,
Rich