In how many ways can the digits 1 through 9 be arranged such that

[Polymath89]

In how many ways can the digits 1 through 9 be arranged such that

In how many ways can the digits 1 through 9 be arranged such that the even digits appear in ascending order?

Well I don't really have a good idea on how to solve this.

If we start with the scenario:

2 4 6 8 _ _ _ _ _ there are 5! arrangements

We could also shift the 2 4 6 8 block five times and still have them in ascending order.

Also we could shift the 8 in 6 ways, so that:

2 4 6 remains fixed and we would have 6! ways for the other numbers to be arranged, but we would count the scenario with 8 in the fourth place twice.

After that it gets a little more difficult. If I fix the 2 and 4 in the first two places, how many arrangements can I make so that the 6 is in front of the 8?

Also is there an easier way to solve this?

 

[vela]

You have to choose 4 spots out of the 9 in which to place the even digits. How many ways can you do that?

 

[Polymath89]

I think 9!/5!, but wouldn't that ignore the fact that the odd integers can be arranged too?

 

[vela]

9!/5! would indeed ignore the fact that the odd integers can be arranged; it also takes into account the even integers can be permuted, which isn't what you want. In other words, you'd be counting, for example, 246813579 and 248613579 separately even though the second arrangement isn't allowed because the even digits aren't ascending.

9P4 = 9!/5! is the number of permutations. You want the number of combinations, 9C4, because you know you'll be filling those four spots in ascending order.

An alternative, and probably easier, way of looking at it is to focus on the odd digits. Figure out the number of ways to pick 5 spots for the odd digits. The remaining four spots will be filled with the even digits in ascending order.