Question

find the number of positive integers which can be formed by using any number of digits from 012345 but using each digit no more than once in each number. how many integer are greater than 3000​

Answer 1

Let us do this step by step!

First, we will find 1-digit numbers

Since the number must be a positive interger, we exclude 0 from our answer. So we have 5 1-digit numbers.

Now for 2-digit mumbers   _ _

Since it must be a 2-digit number, we cannot include 0 infront, so we can choose only 5 numbers for the first space.

Now we can choose 0 for the second space but recall that one number is already used up, so we have 5 possibilities for the second space.

Number of 2-digit numbers is 5×5=25

Now for 3 digit numbers we are going to use the same principle

_ _ _

5×5×4=100

Same for 4-digit numbers: 5×5×4×3=300

5-digit numbers: 5×5×4×3×2=600

And 6-digit numbers: 5×5×4×3×2×1=600

Total number of +ve integers that can be formed is 5+25+100+300+600+600=1630

Now we must see how many of those are greater than 3000

For 5-digit and 6-digit numbers, this is automatically true. But we must see for 4-digit numbers.

We have 4 “spaces” _ _ _ _

For the number to be greater than 3000, the first space (thousand) must be 5, 4 or 3 but cannot be 2, 1 or 0

So we have 4 possibilities for the first space, and the rest follows accordingly

3×5×4×3=180

So total number of +ve integers >3000 is 600+600+180=1380

I hope this helped you! :)