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Question

In how many ways can the letters of the word $$PERMUTATIONS$$ be arranged, if the words start with $$P$$ and end with $$S$$?


A
1814400
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B
1814405
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C
1824050
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D
None of these
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Solution

The correct option is A $$1814400$$
Let first position be $$P$$ and last position be $$S$$(both are fixed)
Since letter are repeating
Hence we use this formula $$\dfrac{n!}{{p}_{1}!{p}_{2}!{p}_{3}!}$$
Total number of letters$$=n=10$$
and since, $$2T$$
$$\Rightarrow {p}_{1}=2$$
Total arrangements$$=\dfrac{10!}{2!}=181440$$

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