In how many ways can the letters of the word PERMUTATIONS be arran | Filo

Class 11

Math

Algebra

Permutations and Combinations

583
150

In how many ways can the letters of the word be arranged if the.i) Words start with and end with .ii) vowels are all togetheriii) there are always letters between and

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

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.

Number of vowels in the given word are $$5$$.

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 .

Here, $$P$$ and $$S$$ are fixed. These $$P$$ and $$S$$ can interchange their position. Hence, $$P$$ and $$S$$ can be arranged in $$2!$$ ways.
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 $$=2!\cfrac {^{10}C_4}{2!}\, 4! \, 7!$$

$$=25401600$$
583
150

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