Class 11

Math

Algebra

Permutations and Combinations

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 togetheriii) there are always $4$ letters between $P$ and $S$

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.

These ten letters with $$T$$ occurring twice can be arranged in $$\cfrac{10!}{2!}$$.

Hence, required number of arrangements =$$\cfrac {10!}{2!}=1814400$$

ii) There are total $$12$$ letters in the word $$PERMUTATIONS$$, with $$T$$ repeated twice.

Since vowels have to always occur together, so they are considered as a single object .

This single object (letter) together with the remaining $$7$$ letters will give us $$8$$ objects (letters).

These $$8$$ objects in which there are $$2$$ $$T's$$ can be arranged in $$\cfrac{8!}{2!}$$ 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 $$=\cfrac {8!}{2!} \times 5!= 2419200$$

iii) There are $$12$$ letters in word $$PERMUTATIONS$$. $$T$$ occurs twice .

Since, $$P$$ and $$S$$ are fixed, there are $$10$$ letters left , where $$T$$ occurs twice .

From these $$10$$ letters, $$4$$ letters can be chosen in $$\cfrac{^{10}C_4}{2!}$$ ways.

Now since, $$4$$ letters are between $$P$$ and $$S$$, so these six letters can be considered as a single object (letter). These four letter (between $$P$$ and $$S$$ )can be arranged in $$4!$$ ways.

Now, the remaining $$6$$ letters and 1 object i.e. total $$7$$ can be arranged in $$7!$$ ways.

Hence, required number of ways $$=2!\cfrac {^{10}C_4}{2!}\, 4! \, 7!$$

$$=25401600$$