The number of integers lying between 3000 and 8000 (including 3000 and 8000) which have at least two digits equal is
(a) 2481
(b) 1977
(c)4384
(d) 2755

Dear student,
The total number of numbers between 3000 and 8000 (including both of them)is 8000–3000+1 = 5001.
 
Now for the numbers with no digits equal, we consider numbers 3000–7999 (since 8000 already violates our condition, so we don’t have to treat it separately).
 
The thousands place can be filled using any of the digits 3 to 7. So that is 5 ways. Now, since the numbers can’t repeat, the hundreds place, tens place, units place can be filled by 9 ways (0 to 9 is 10 ways minus the number used in the thousands place amounts to 9 ways), 8 ways and 7 ways respectively (since the number already used can’t be used, the number of ways of filling any place keeps on decreasing).
 
So the number of numbers where no digit repeats is
 
5∗9∗8∗7=2520 numbers.
 
Thus the answer we want is 5001−2520=2481 numbers.
Option (a) is correct
Regards

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