Question

How many numbers are there in all from 4000 to 4999 (both 4000 and 4999 included) having at least one of their digits repeated

Answer 1

Hi.....✌


There are 1000 numbers from 4000-4999 (inclusive )



◆ 1000 - (9×8×7) = 496

Answer 2

Answer:

There are 496 numbers were there.

Step-by-step explanation:

Given : Numbers are there in all from 4000 to 4999 (both 4000 and 4999 included) having at least one of their digits repeated.

To find : How many numbers are there.

Solution :

There are 1000 number from 4000–4999 both inclusive.

Since, 4 is already fixed in thousands position, we have 9 possibilities (0–9) in hundreds. As we cannot use the same number again, we have 8 possibilities in tenths position and similarly 7 possibilities in units.

So, The unique number would be $9\times 8\times 7$

Numbers will be  $9\times 8\times 7=504$

Total number is $1000-504=496$

Therefore, There are 496 numbers were there.