Question

Question no : 2

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Question : The number of integers lying between 3000 and 8000 (including 3000 and 8000) which have at least two digits equal?

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Answer 1

First, we can ignore 8000, since it doesn't have all digits different, so we can just go up to 7999. To have all 4 digits different, there will be 5 choices for the first digit (3 through 7), 9 for the second digit (0 through 9, minus whatever the first digit was), 8 for the third (0-9 minus the first two digits), and 7 for the 4th, so a total of 5*9*8*7 = 2520 choices. 
All the rest will have at least 2 digits in common. Since there are 5001 numbers in question, this leaves 2481 with at least 2 digits in common. 

Answer 2

including 3000 and 8000 there are 4 integers which have at least 2 digits equals which are 2481, 1977,4384, 2755
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