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16 Sep 2014, 01:53
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E
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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
A. 9
B. 12
C. 36
D. 72
E. 108
16 Sep 2014, 01:53
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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:
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 \(3 \times 2 \times 3 \times 2 \times 3 = 108\).
Answer: E
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 \(3 \times 2 \times 3 \times 2 \times 3 = 108\).
Answer: E
07 Jul 2015, 11:23
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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.
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.
