# Counting techniques problem

How many numbers greater than $50,000,000$ can be formed by rearranging the digits of the number $13,979,397$?

Can someone give me a hint?

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Going by the hint given by one of the responder:

It is 7(1-1s,3-9s,2-3s, 1-7s) = Total number of arrangements of 7 remaining numbers is = $$\frac{7!}{2!3!} = 420$$

It is 9(1-1s,2-9s,2-3s,2-7s) = Total number of arrangements of 7 remaining numbers is = $$\frac{7!}{2!2!2!} = 630$$

Sum it up to the answer of 1050.

Thanks

Satish

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Hint: the rearrangements that are greater than $50,000,000$ are exactly those whose first digit is $7$ or $9$.

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