**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,

(ii) there are always 4 letters between P and S?

**Solution:**

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 $=\frac{8 !}{2 !} \times 5 !=2419200$

(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 × 7 = 14 ways.

Therefore, by multiplication principle, required number of arrangements in this case $=\frac{10 !}{2 !} \times 14=25401600$

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