In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end wi

Exercise 6.3 · Q9

Q. 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) there are 5 letters between P and S

(iii) vowels are all together

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PERMUTATIONS $=$ P,E,R,M,U,T,A,T,I,O,N,S (12 letters, T repeated twice). Handle the repeated letter with $\dfrac{n!}{p!}$ throughout.

n

(i) Words start with P and end with S

Fix P in position 1 and S in position 12 (1 way each, since both are non-repeated letters).
The remaining 10 letters (E,R,M,U,T,A,T,I,O,N — includes T twice) fill the 10 middle positions: $\dfrac{10!}{2!}$ .
Compute: $10!=3628800$ ; $3628800/2=1814400$ .

(ii) There are 5 letters between P and S

"5 letters between P and S" means their positions differ by $6$ (e.g. position $i$ and $i+6$ , so the letters at $i+1,\ldots,i+5$ — 5 of them — sit between).
In a 12-letter word (positions 1–12), position pairs with difference $6$ : $(1,7),(2,8),(3,9),(4,10),(5,11),(6,12)$ — 6 pairs.
For each pair, P and S can occupy either position (P first & S second, or S first & P second): $\times2$ orders. So placements of P,S $=6\times2=12$ . …

Q.In how many ways can the letters of the word PERMUTATIONS be arranged if the