Question

In how many ways can a letters of the word $PERMUTATIONS$ be arranged if the
(i) Words with start with $P$ and end with $S$
(ii) vowels are all together
(iii) There are always $4$ letter between $P$ and $S$

Answer 1

$\left(i\right)$ Given word is PERMUTATIONS

Now we have to find no. of ways we can arrange the letters with starting $P$ and end with $S$

Total letters =$10$ in wj=hich we have $2T^{\prime }s$,

if $P$ and $S$ are fixed

then no. of arrangement $=\frac{10!}{2!}=181440$

$\left(ii\right)$ vowels in given word are : $AEIOU$

considering vowels as one string

Then $AEIOU$ PRMTTNC

can be arrange in $\frac{-8}{2}\times 5!=2419200$

$\left(iii\right)$ we have to select $4$ letters between $P$ and $S$ and their are $10$ letters

Therefore No. of ways ${=}^{10}{C}_2=210$

and , $P$ and $S$ can also change their position

$\therefore$ No of ways $=420$ [Double of $210$]

Now $4$ letters can arrange in $4!$ ways

Now no of ways four letters arrange between $P$ and $S$ are $4!\times 420=10080$

Now consider $P4S$ as one string

Now remaining 6 letters +one string can

be arrange in $7!$ ways

Total no. of ways $=10080\times 7!/2$

$=\frac{50,803,200}{2}=25401600$