4 digits no that can form =3*5*4*3=180
since it's greater than 3000..on 1st place there will be 3,4 or 5
5 digits no =5*5*4*3*2=600
bcoz 0 can't come on 1st place..n similarly
6 digit no=5*5*4*3*2*1=600..
..so total..600+600+180=1380..
..i hope it's correct..

Counting numbers greater than 3000 using the digits 0, 1, 2, 3, 4, and 5 without repetition

To count the numbers greater than 3000 using the given digits without repetition, we need to consider the possible combinations of these digits and exclude any combinations that result in a number less than or equal to 3000.

Step 1: Counting numbers without repetition
To determine the total number of possible combinations without repetition, we can use the concept of permutations. Since the digits can be arranged in a specific order, we can calculate the number of permutations using the formula:

nPr = n! / (n - r)!

Where n is the total number of items and r is the number of items selected at a time.

In this case, we have 6 digits (0, 1, 2, 3, 4, 5) and we need to select 4 digits to form a number greater than 3000. Therefore, the number of possible combinations without repetition is:

6P4 = 6! / (6 - 4)! = 6! / 2! = 6 x 5 x 4 x 3 = 360

Step 2: Excluding numbers less than or equal to 3000
To exclude numbers less than or equal to 3000, we need to consider the position of the thousands digit. Since the number needs to be greater than 3000, the thousands digit must be either 3, 4, or 5.

For each of these three possibilities, we can calculate the number of valid combinations by multiplying it with the number of possible combinations for the remaining three digits.

For the thousands digit as 3:
- The remaining three digits can be selected from the available digits (1, 2, 4, 5) in 4P3 = 4! / (4 - 3)! = 4! = 4 x 3 x 2 = 24 ways.

For the thousands digit as 4:
- The remaining three digits can be selected from the available digits (0, 1, 2, 5) in 4P3 = 4! / (4 - 3)! = 4! = 4 x 3 x 2 = 24 ways.

For the thousands digit as 5:
- The remaining three digits can be selected from the available digits (0, 1, 2, 3, 4) in 5P3 = 5! / (5 - 3)! = 5! / 2! = 5 x 4 x 3 = 60 ways.

Step 3: Calculating the total number of valid combinations
To calculate the total number of valid combinations, we sum up the number of valid combinations for each thousands digit:

24 + 24 + 60 = 108

Therefore, there are 108 numbers that can be made using the digits 0, 1, 2, 3, 4, and 5, which are greater than 3000 without repetition.