ANSWER
The word "BULLET" has 240 possible permutations such that the vowels are never together.
EXPLANATION
Step 1: Count how many possible arrangements will a six-digit word with two vowels have such that the vowels are not together.
Let us represent the vowels as "V" and the consonants as "C".
VCVCCC - 1
VCCVCC - 2
VCCCVC - 3
VCCCCV - 4
CVCVCC - 5
CVCCVC - 6
CVCCCV - 7
CCVCVC - 8
CCVCCV - 9
CCCVCV - 10
There are 10 possible arrangements so that the vowels are not together.
Step 2: Find how many possible arrangements do the vowels and consonants of the word "BULLET" have.
Vowels = EU = 2 letters = 2!
↓
Vowels = 2 possible arrangements
[tex]Consonants = BLLT = \frac{4 \: letters}{2 \: repeated \: } = \frac{4!}{2!} [/tex]
[tex] \frac{4!}{2!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1} = 4 \times 3 = 12[/tex]
↓
Consonants = 12 possible arrangements
Step 3: Multiply all the answers you got.
10 × 2 × 12 = 240
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