In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i). Words start with P and end with S,
(ii). Vowels are all together,
(iii). There are always $4$ letters between P and S?
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Ans:
In the word PERMUTATIONS, there are $2$ T’s, and all the other letters appear only once.
If P and S are fixed at the extreme ends (P at the left end and S at the right end), then $10$ letters are left.
Hence, in this case, required number of arrangements
$\frac{10!}{2!}=1814400$.
Ans:
There are $5$ vowels in the given word, each appearing only once.
Since they have to always occur together, they are treated as a single object for the time being. This single object together with the remaining $7$ objects will account for $8$ objects, in total.
These $8$ objects in which there are $2$ T’s can be arranged in $\frac{8!}{2!}$ ways.
Corresponding to each of these arrangements, the $5$ different vowels can be arranged in $5!$ ways. Therefore, by multiplication principle, required number of arrangements in this case
$\frac{8!}{2!}\times 5!=2419200$.
Ans:
The letters have to be arranged in such a way that there are always $4$ letters between P and S. Therefore, in a way, the places of P and S are fixed. The remaining $10$ letters in which there are $2$ T’s can be arranged in $\frac{10!}{2!}$ ways.
Since,Possible places of P & S are 1 & 6, 2 & 7, 3 & 8, 4 & 9, 5 & 10, 6 & 11, 7 & 12. Then there can be $4$ letters between P & S.Also P & S can be interchanged as it won’t effect the number of letters between them.
So, the letters P and S can be placed such that there are $4$ letters between them in $2\times 7=14$ ways.Therefore, by multiplication principle, required number of arrangements in this case
$\frac{10!}{2!}\times 14=25401600$.