Question

How many 5 digit numbers are there such tat 2 left most digits are even and remaining ate odd?

Answer 1

Answer:

Think of the five places of the five digit number as boxes.

According to the given condition, the first box(from the left) can be filled by any of the four digits namely 2,4,6,8 but not 0, as in that case it would become a four digit number. So there are 4 ways possible to fill the box.

Now for the second box there are 5 even numbers (available) that can be filled, hence we have 5 ways.

For the test three boxes too, we have 5 odd numbers , hence 5×5×5 ways to fill the three boxes.

So total number of ways to fill the five boxes is

4×5×5×5×5 that is 4×5^4 ways.

So total possible five digit number is 4×5^4, that is 2500.

Hope this will be helpful for you.

Answer 2

Answer:

2500

Step-by-step explanation:

Just list down all.