Question

How many words (with or without meaning) can be formed from the letters of the word, 'DAUGHTER', so that

(i) all vowels occur together?

(ii) all vowels do not occur together?

Answer 1

The given word, 'DAUGHTER' contains 3 vowels A, U, E and 5 consonants, D, G, H, T, R.

Case (i) When all vowels occur together :

Let us assume (AUE) as a single letter.

Then, this letter (AUE) along with 5 other letters can be arranged in ${}^6{P}_6=(6!)$ ways = $(6\times 5\times 4\times 3\times 2\times 1)$ ways

= 720 ways.

These 3 vowels may be arranged among themselves in $3!=6$ ways.

Hence, the required number of words with vowels together

= $(6!)\times (3!)=(720\times 6)=4320.$

Case (ii) When all vowels do not occur together.

Number of words formed by using all the 8 letters of the given word

= ${}^8{P}_8=8!=(8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)=40320.$

Number of words in which all vowels are never together = (total number of words) - (number of words with all vowels together)

=$(40320-4320)=36000.$