Permutation and combination

[lone monolith]

How many numbers with 6 digits can be formed, taking digits from 1 to 9, any number of times, so that the digits in the first three places are not present in the last three places?

[harsh python]

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[lone monolith]

I thought I got my head around this but that didn't last long

I don't understand the purpose of multiplying by 2

case 2 is just a,a,b right? where only 1 number is different

* 3!/2! is just to account for the 2 repeated numbers

the 3! is hard to understand, why are we even multiplying that, is it because 3! is is the total combinations we can do with the 3 digits?

same for case 3, i understand 9C3 * 3! and then 6C1 because we now have 6 numbers to choose as we've already taken 3 of them

but the last one where you multiply 6C1 * 2 I don't get

I'd appreciate the help if someone helps me rid my confusions

[oak palm]

Yea sort of

you can mix around 3 digits 3! times

9C2 only gives the choices of 2 digits

[oak palm]

the first case doesn't have 3! because you can't exaclty mix around the same digits

Not sure why they multiplied by 2 though

Because if you swap the first 3 blocks and the last 3 blocks you're overcounting

[lone monolith]

ye that has me stumped

[glossy ibex]

are we supposed to consider this? or are you okay with me starting over

[lone monolith]

i'm completely fine with you starting over I sort of understood it with different approaches myself, just case 3 is a little confusing

[glossy ibex]

Ah case 3, I take it's the case where nothing repeats in either the first three or last three spots

Should it not be the number of permutations you can come up with using 6 numbers from these 9?

[oak palm]

I think it means last 3 numbers can repeat, only the first 3 digits don't repeat

[lone monolith]

ig it does make sense, we choose 9C3 * 3! becasue there are no repeats, we chose 3 unique numbers from 9 then we needed to pick 6 remaining

again, the only thing i don't get is the * 2

and *3! because order does matter, it's a permutation

but *2?

[glossy ibex]

Ah i interpreted it as no repeats anywhere

[glossy ibex]

Maybe case 3 is : no repeats in first 3 OR last 3, but that runs into over counting issues

[lone monolith]

i dont think its that complicated

[harsh python]

@lone monolith

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