SOLUTION: If the letters word'DAUGHTER'are to be arranged so that vowels occupy the odd places, then the number of different words are? (Solved by only permutations )

Algebra ->  Permutations -> SOLUTION: If the letters word'DAUGHTER'are to be arranged so that vowels occupy the odd places, then the number of different words are? (Solved by only permutations )      Log On


   



Question 1055924: If the letters word'DAUGHTER'are to be arranged so that vowels occupy the odd places, then the number
of different words are? (Solved by only permutations )

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!


D A U G H T E R
1 2 3 4 5 6 7 8

[Notice that DAUGHTER itself is NOT one of the 
required permutations because A is in position 2,
which is not an odd position]

There are 4 odd positions: 1,3,5,7.

There are 3 vowels: A,E,U

An example of one of the desired permutations would 
be RTEHAGUD, 

R T E H A G U D
1 2 3 4 5 6 7 8

Take away the consonants and leave only the
vowels for illustration:

    E   A   U  
1 2 3 4 5 6 7 8

The permutation of odd positions chosen here
for A,E,U were 5,3,7 respectively.

We placed the A in position 5, the E in position 3, 
and the U in position 5.

    E   A   U  
1 2 3 4 5 6 7 8

There are 4P3 permutations of 3 of the odd positions 1,3,5,7

The one that was chosen in this example was 5,3,7.

In that example, We placed A in the odd position 5 that 
came first in the PERMUTATION 5,3,7 of odd positions,
(not that came first in the word!).  

Then we placed E in the odd position that came second 
in the PERMUTATION 5,3,7 of odd positions, (not that 
came second in the word!).  

Finally we placed U in the odd position that came 
third in the PERMUTATION 5,3,7 of odd positions, (not that 
came third in the word!)

There are 5 remaining positions in which to place the 5 
remaining letters which is 5P5.

Answer: (4P3)(5P5) = 24(120) = 2880.

Edwin