. In how many ways can the letters of the word

'PERMUTATIONS' be arranged if the

(i) words start with P and ends with S and 

(ii) Vowels are all together.


Best Answer

Explanation: In the word PERMUTATIONS, there are two T's  and all the other letters appear only once.

(i) If p and s are fixed at the extreme ends, then 10 letters are left. 

p_ _ _ _ _ _ s

10!/2! =10×9×8×7×6×5×4×3×2!  / 2!

= 1814400

So , in this case, the required number of arrangements

(ii) There are five vowels in the given word, each appearing only once.

Since they have to always occur together, they are treated as a single object.

So, a single object together with the remaining 7 objects will be 8 objects. 

These 8 objects in which there are two  T's  can be arranged in 8! / 2!

Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5! ways. 

Therefore, the required number of arrangements 

8! /2!× 5! =8×7×6×5×4×3×2! / 2! × 120

= 2419200

Final Answer: (i) The required number of arrangements when  T' and S' are fixed at the extreme ends is 1814400.

(ii) The required number of arrangements when all vowels are together is 2419200.


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