Question

Find the number of different $8-$ letter arrangements that can be made from the letters of the word $DAUGHTER$ so that:
$\left(i\right)$ All vowels occur together
$\left(ii\right)$ All vowels do not occur together.

Answer 1

$\left(i\right)$ All vowels occur together.
Total number of letters in $DAUGHTER=8$
number of vowels in $DAUGHTER=3$
$\because$ All vowels occur together, so assume $AUE$ as single letter.
$\therefore$ Letters become $AUE,D,G,H,T,R\to 6$ letters.

Arranging $3$ vowels.
$\therefore$ Number of Permutations of $3$ vowels $={}^3{P}_3=6$ ways

Number of Permutation of $6$ letters.
${=}^6{P}_6$
$=\frac{6!}{(6-6)!}=\frac{6!}{0!}=720$

Finding total number of arrangements.
Total number of arrangements.
$=720\times 6=4320$.

$\left(ii\right)$ All vowels do not occur together.
Number of letters in $DAUGHTER=8$
Total number of permutations ${=}^8{P}_8$
$=\frac{8!}{(8-8)!}=\frac{8!}{0!}$
$=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$
$=40320$
Number of permutations in which all vowels never together.
$=$ Total number of permutations $-$ Number of permutations in which all vowels come together
$=40320-4320$.
$=36000$.