Question

Find the number of ways of permuting the letters of the word PICTURE so that
(i) All vowels come together
(ii) No two vowels come together
(iii) The relative positions of vowels and consonants are not disturbed

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:
The letter PICTURE has $3$ vowels $\{ E,I,U\} $and $4$ consonants $\{ C,P,R,T\} $
(i) Let us suppose $3$ vowels as one unit and the number of permutations in $E,I,U$ is$3!$.
Now, we can arrange $4$ consonants $ + 1$units of vowels in P,I,C,R,EIU $ = 5$ ways.
So, permutations will be $5!$.
Hence, the number of permutations on which $3$ vowels occur together \[ = 5! \times 3!\]
$ = 5 \times 4 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1$
$ = 120 \times 6$
$ = 720$
(ii) No two vowels occur together. We arrange $4$ consonants $\{ C,P,R,T\} $. Permutation of $4$ consonants is $4!$ and permutation of vowels EIU is $3!$
There are $5$ letters and $3$ vowels, so we can write as $^5{C_3}$.
$\therefore $The number of words in which no two vowels occur together$ = 4!\, \times {\,^5}{P_3}$
$ = 4 \times 3 \times 2 \times 1 \times \dfrac{{5!}}{{(5 - 3)!}}$
$ = 4 \times 3 \times 2 \times 1 \times \dfrac{{5 \times 4 \times 3 \times 2!}}{{2!}}$
$ = 24 \times 60$
$ = 1440$
(iii) The relative position is not altered: we arrange $3$ vowels in $3!$ ways.
We arrange $4$ consonants in $4!$ ways
It gives the total permutation$ = 4!\, \times 3!$
$ = 4 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1$
$
   = 24 \times 6 \\
   = 134 \\
 $


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