SOLUTION: In a random arrangement of letters of word VIOLENT how many ways are there that the vowels occupy odd positions only

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Question 360325: In a random arrangement of letters of word VIOLENT how many ways are there that the vowels occupy odd positions only
Found 2 solutions by Theo, Edwin McCravy:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
without any restrictions, the number of possible ways to arrange the letters VIOLENT would be 7 * 6 * 5 * 4 * 3 * 2 * 1 = 7! = 5040

this can only happen in one way.

this assumes that each letter can only be used once.

since there are 3 vowels and 4 consonants, and the vowels can only be in odd positions, then the possible ways this can happen are provided by the following formula:

number of possible ways to get sets of 3 out of a total of 4 is:

4! / (3! * 1!) = (4*3!)/3! = 4

the general equation for this is:

n! / (x! * (n-x)!)

n = 4
x = 3
(n-x) = 1

the number of combinations in each one of these ways is equal to 4! * 3! = 144

since there are 4 ways that can happen as we just calculated above, then the total ways you can get 3 vowels in only the odd position is 4 * 144 = 576.

the 4 possible positions the vowels can be in are shown as follows:

x shows the position of the vowel.
c shows the position of the consonant. 


position          1   2   3   4   5   6   7
way number 1      x   c   x   c   x   c   c
way number 2      x   c   x   c   c   c   x
way number 3      x   c   c   c   x   c   x
way number 4      c   c   x   c   x   c   x


the number of possible combinations in each of the ways is calculated as follows:

we'll take way number 1 to show you.

the other ways work the same.

way number 1

vowel in 1st and 3d and 5th position.

you have 3 vowels.

the first vowel position can take any 3 of them.
the second vowel position can take any 2 of them.
the third vowel position can take any 1 of them.

total possible combinations are therefore 3*2*1 = 6 = 3! for the vowels.

you have 4 consonants.

the first consonant position can take any 4 of them.
the second consonant position can take any 3 of them.
the third consonant position can take any 2 of them.
the fourth consonant position can take any 1 of them.

total possible combinations are therefore 4*3*2*1 = 24 = 4! for the consonants.

for each possible combination of vowels, there are 24 possible combinations of consonants.

since there are 6 possible combinations of vowels, then the total number of possible combinations of vowels and consonants for way number 1 is 6 * 24 = 144.

4 possible ways means 4 * 144 = 576.












Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!

The other tutor made it unnecessarily complicated.

There are 3 vowels: I, O, E 

and 

there are 4 consonants V, L, N, T

and

there are 4 odd positions: 1st letter, 3rd letter, 5th letter, 7th letter.

So we choose 3 of the 4 odd positions for the vowels in 4C3 or 4 ways.
Having chosen which of the 3 odd positions the 3 vowels are to go in, 
we can then arrange the three vowels in those odd positions in 3P3 
or 3! ways.  
Having placed the vowels, the other 4 positions are filled with the 
4 consonants in 4P4 or 4! ways.

That's 4*3!*4! = 576

Edwin
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