Question

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?

Answer 1

In the word PERMUTATIONS, there are 2 Ts and all the other letters appear only once.

(i) 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 = $\frac{10!}{2!}=1814400$

(ii) 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. These 8 objects in which there are 2 Ts 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 = $\frac{8!}{2!\times5!}=2419200$

(iii) 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 Ts can be arranged in $\frac{10!}{2!}$ ways
Also, 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 = $\frac{10!}{2!\times14}=25401600$