How many sequences $a_1, a_2, a_3, a_4, a_5$ satisfying $a_1 < a_2 < a_3 < a_4 < a_5$ can be formed if each $a_i\in\{ 1, 2, 3, …, 9\}$

Question

How many sequences $a_1, a_2, a_3, a_4, a_5$ satisfying $a_1 < a_2 < a_3 < a_4 < a_5$ can be formed if each $a_i$ must be chosen from the set $\{1, 2, 3, …, 9\}$?

What I can determine so far: Order matters, making this a permutation. $a_1$ cannot be a number larger than $5$, otherwise there would not be enough larger numbers to fulfill the condition that $a_1 < a_2 < a_3 < a_4 < a_5$. Does this imply 5 individual cases? At the same time, there needs to be a number to fill in each space. I'm stumped.

Answer 1

HINT

Notice that for any $5$ different numbers there is one and only way to put them into a valid sequence.

So, the number of possible sequences simply equals the number of ways you can pick $5$ numbers out of the $9$ numbers, ... which I am sure you can figure out.

Answer 2

In order to simplify this problem, instead of picking $a_1,a_2,...,a_5$, we can pick $a_1,a_2-a_1,a_3-a_2,...,a_5-a_4$. Why is this helpful? First, because ensuring that each of these numbers are positive enforces the inequality you've presented. Second, if we label each element of our new sequence $v,w,x,y,z$, we know that $$v+w+x+y+z=9$$Now, we can use stars and bars. We have to distribute 9 "units" among 5 numbers, ensuring that each one is at least one. This leads us to the combination of $${9\choose5}=126$$

Another approach is to consider that each unique set of 5 numbers I pick can be ordered such that $a_1<a_2<...<a_5$. As such, this just becomes a problem of selecting 5 numbers from $\{1,...,9\}$, giving us $9\choose5$.

Answer 3

With each subset of 5 element in this set uniqely determine such sequence, so the answer is $${9\choose 5}$$