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How many numbers are there between 100 and 1000 such that every digit is either 7 or 9 ?

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Between 100 and 1000 , all numbers are of there digits
Number of ways of filling hundred's place =2 (`because` either 7 or 9 is to be used)
Number of ways of filling ten's place =2
Number of ways of filling units place =2
`therefore ` By the principle of multiplication total number of `=2xx2xx2=8`
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