How many five digit number can be formed such that their numbers counted from left to right creates a decreasing sequence?

Question

How many five digit number can be formed such that their numbers counted from left to right creates a decreasing sequence?

Numbers $= {0,1,2,3,4,5,6,7,8,9}$

Example: $54321$ and $96310$

If we would count the amount of five digit numbers that can be formed such that their number counted from left to right is INCREASING we would get:

$\binom{9}{5}=126$

Since the amount of distinct set can be sorted so $n_1>n_2>n_3>n_4>n_5$ Were $n_1$ is the digit with the highest value.

My attempt at the real question in hand is the following:

We sort every digit so that $n_1<n_2<n_3<n_4<n_5$

We would now get $\binom{10}{5}=252$ decreasing five digit number. This is clearly wrong since I don't account for the zero. I need to divide this problem into different parts in some way, any got any suggestions? Thanks beforehand!

Answer 1

Hint the answer will be ${10\choose 5} $ because when you select any $5$ digits there will always be a sequence st all the digits form a decreasing order.

Answer 2

The question can be interpreted in another way:

How many decreasing functions from $\mathbb{N}_{10} \rightarrow \mathbb{N}_{10}?$

And the answer to this question is clearly $\binom{10}{5}$ because you are choosing $5$ numbers, without repetition, in a set of $10$ numbers. So, in any way you choose them, there is always going to be a decreasing set.

Answer 3

You only need a group of 5 numbers out of 10 numbers, without repetition which would be as you have calculated, 252. Later then, you can order these numbers in decreasing order to form the desired numbers.