Question

Find the number of arrangements of the letters of the word $INDEPENDENCE$. In how many of these arrangements,
$\left(i\right)$ do the words start with $P$
$\left(ii\right)$ do all the vowels always occur together
$\left(iii\right)$ do the vowels never occur together
$\left(iv\right)$ do the words begin with $I$ and end in $P$ ?

Answer 1

In word $INDEPENDENCE,$
$3Ns,4Es,2Ds$ and $1I,1P$ and $1C$ are there (repetition)
$n=12,P_1=3,P_2=4$ and $P_3=2$
$\therefore$ Total arrangements $=\frac{n!}{P_1!P_2!P_3!}$
$=\frac{12!}{3!4!2!}$
$=\frac{12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!\times 2\times 1}$
$=1663200$

$\left(i\right)$ words start with $P$

$P---------$
So, $4Es,3Ns,2Ds$ (repetition) are there and $n=11$ letters
$\therefore$ number of arrangements
$=\frac{11!}{4!3!2!}=138600$

$\left(ii\right)$ Arranging vowels.

Total number of vowels $\left(n\right)=5$
Repetition $\to 4Es$ and $1I$
$\therefore n=5$ and $P_1=4$
Total arrangement $=\frac{5!}{4!}$

Arranging the letters when vowels are taken as one unit.
Total number $=n=8$
Here, $3Ns,2Ds\to$ repetition
Total arrangements $=\frac{8!}{3!2!}$
$\therefore$ Required number of arrangements
$=\frac{8!}{3!\times 2!}\times \frac{5!}{4!}=16800$

$\left(iii\right)$ Number of arrangements when vowels never occur together $=$ total number of arrangements (without any restriction) $-$ the number of arrangements where all the vowels occur together
$=1663200-16800=1646400$

$\left(iv\right)$ Let $I$ and $P$ fix at extreme ends.

$I----------P$

$10$ letters in which $2D,4E$ and $3N\to$ repetition
So, $n=10,P_1=2,P_2=4$ and $P_3=3$
$\therefore$ Required number of arrangements
$=\frac{10!}{2!4!3!}=12600$.