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

Total letters in the word PERMUTATIONS = 12.

Here T = 2.

(i) Now first letter is P and last letter is S, which are fixed.

So the remaining 10 letters are to be arranged between P and S.

$\therefore$ Number of permutations = $\frac{10!}{2!}$

= $\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2!}{2!}$

= $1814400$

(ii) There are 5 vowels in the word PERMUTATIONS. All vowels can be put together.

$\therefore$ Number of permutations of all vowels together = ${}^5{P}_5$

= $\frac{5!}{0!}=5\times 4\times 3\times 2\times 1=120$

Now consider the 5 vowels together as one letter. So the number of letters in the word when all vowels are together = 8.

$\therefore$ Number of permutations = $\frac{8!}{2!}$

= $\frac{8\times 7\times 6\times 5\times 4\times 3\times 2!}{2!}$

= $20160.$

Hence the total number of permutations
= $120\times 20160=2419200$

(iii) Here P and S are on 1st and 6th places, P and S are on 2nd and 7th places, P and S are on 3rd and 8th places, P and S are on 4th and 9th places, P and S are on 5th and 10th places, P and S are on 6th and 11th places, P and S are on 7th and 12th places. Now we see that P and S can be put in 7 ways and also P and S can interchange their positions.

$\therefore$ Number of permutations = $2\times 7=14$

Now the remaining 10 places can be filled with remaining 10 letters.

$\therefore$ Number of permutations = $\frac{10!}{2!}$

= $\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2!}{2!}$

= $1814400$

Thus total number of permutations = $14\times 1814400=25401600$