Question

Find the number of different $8 - $ letter arrangements that can be made from the letters of the word “DAUGHTER” such that all vowels do not occur together.
A. $36000$
B. $4320$
C. $10080$
D. $40230$

Answer 1

Hint: Permutation of a set is an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its element.
Number of permutation $ = $total number of permutation – number of permutation of all vowels occurring together


Complete step by step solution:
Hint: Permutation of a set is an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its element.
Number of permutation $ = $total number of permutation – number of permutation of all vowels occurring together
Complete step by step solution:
Total number of letters in DAUGHTER$ = 8$
Total number of permutation of $8$letters $ = {\,^8}{P_8}$
$ = \dfrac{{8!}}{{(8 - 8)!}}$
$ = \dfrac{{8!}}{{0!}}$
$ = \dfrac{{8!}}{1}$
$ = 8!$
$ = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$ = 40320$.
And vowels in DAUGHTER$ = A,V$and $E$.
Since all vowels occur together, assume $AUE$ as a single object and left letters are DGHTR
Arrange $3$vowels $ = {\,^3}{P_3}$
                                                                           Arrange $6$letters$ = 5 + 1 = 6$
$ = \dfrac{{3!}}{{3 - 3!}}$
                                                                           $ = {\,^6}{P_6}$
$ = \dfrac{{3!}}{{0!}}$
                                                                          $ = \dfrac{{6!}}{{6 - 6!}}$
$ = \dfrac{{3!}}{1}$
                                                                          $ = \dfrac{{6!}}{{0!}}$
$ = 3!$
                                                                          $ = 6!$
$
   = 3 \times 2 \times 1 \\
 = 6 \\

                                                                                                          = 6 \times 5 \times 4 \times 3
                                                                                                                \times 2 \times 1
                                                      = 720


 $

So, total number of average $ = 720 \times 6$
$ = 4320$
All vowels do not occur together $ = $total number of permutation – number of permutation of all vowels occurring together.
$
   = 40320 - 4320 \\
   = 36000 \\
 $


Note: Students should solve the value of vowels and consonants separately. Then by multiplying both the values we’ll get the answer.