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Question

How many numbers between $$400$$ and $$1000$$ can be formed with the digits $$0, 2, 3, 4, 5, 6,$$ if no digit is repeated in the same number?


Solution

Number between $$400$$ and $$1000$$ consist of three digits with digit at hundred’s place greater than or equal to $$4$$. Hundred’s place can be filled, by using the digits $$4, 5, 6$$ in $$3$$ ways. Now, ten’s and unit’s places can be filled by the remaining $$5$$ digits in $$^{5}{{P}_{2}}$$ ways.
Hence, the required number of numbers  $$=3\times^{5}{{P}_{2}}=3\times\dfrac{5!}{3!}=3\times\dfrac{5\times 4\times 3!}{3!}=3\times 20=60$$

Mathematics

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