JEE Mathematics 2023 Question with Solution

Given: The word is INDEPENDENCE and all the vowels must always occur together.

Find: The number of arrangements of the letters of the word in which all the vowels occur together.

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

On expanding the factorial, we get

On simplification,

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

After simplification, we get

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.

Therefore, the correct option is A.

Given: All vowels of INDEPENDENCE must stay together.

Find: Total such arrangements.

Treat all vowels as a single block. Then we arrange this vowel block together with the consonants.

The consonants are N, D, P, N, D, N, C and the vowel block acts like one more item. So total items become $8$.

Among these, N repeats $3$ times and D repeats $2$ times. Hence the number of arrangements is

Now arrange the vowels inside the block. The vowels are I, E, E, E, E. Since E repeats $4$ times, the number of internal arrangements is

Therefore, total arrangements are

Hence, the number of arrangements is $16800$.