Question

Let x be the number of integer lying between 2999 and 8001 which have atleast two digit equal . Then what is the value of x.

Answer 1

There are two ways to solve this one;

1. You can check the cases for:

• 2 digits being equal
• 3 digits being equal
• 4 digits being equal

2. Else subtract the number of numbers which have no digits equal and subtract it from the total numbers.

I’ll go with the 2nd way since it’s shorter. The total number of numbers between 3000 and 8000 (inclusive) 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.