Ishel03idn
Answered

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How many different possible
permutations can be made from
the word "BULLET" such that the
vowels are never together
?​

Sagot :

Smart852idn

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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