Question

Consider the word INDEPENDENCE. The number of words such that all the vowels are together is?

• A. 16800
• B. 15800
• C. 17900
• D. 14800

Hint:

To solve this, treat all the vowels (I, E, E, E, E) as a single entity or block. Then, count the arrangements of the consonants and the vowel block together. Afterward, count the different arrangements of the vowels within the block.

Answer 1

The Correct Option is A

The correct answer is (A) : 16800
The word has 12 letters, out of which 5 are vowels and 7 are consonants.
Because we need to keep the vowels together always, we can consider the 5 vowels as one letter.
So, the number of ways of arranging the consonants is given by, \(\frac{8!}{3!×2!}\)
On expanding the factorial, we get , \(\frac{8×7×6×5×4×3!}{3!×2!}\)
On simplification,\(\frac{8×7×6×5×4}{2}\)
=3360
So the vowels can also be rearranged themselves. Out of the 5 vowels, 4 are the same.
So, the ways of arranging the vowels is given by \(\frac{5!}{4!}\)
After simplification, we get, \(\frac{5×4!}{4!}=5\)
Therefore, the vowels can be arranged in 5 ways.
The number of words that can be formed such that vowels are always together is given by the product of the number of ways of arranging the letters with all the vowels together and the number of ways of arranging the vowels.
\( ⇒5×3360=16800\)