Combinatorics, sets

[WinterGraschp]

Into how many distinct, equally sized (8 per group) groups can 32 unique cards be sparated, if 4 of those cards have to be together (eg. in the same group).

My solution was that first we make a set with 4 elements out of 28, because that is the set in which those 4 cards go, then we make a set with 8 elements out of 24, then a set with 8 elements out of 16 and then finally a set of 8 elements out of 8 elements (which is 1). We multiply these so we got something like: (28|4) * (24|8) * (16|8) * (8|8), now this would be correct if the order of the groups would matter, but it doesn't so we divide the whole thing with 4!, because there are 4 groups so we finally get: ((28|4) * (24|8) * (16|8) * (8|8)) / 4!

Now apparently my soulution is fine, except for the fact that we have to divide by 3!, according to the solution.

I'm either right and the solution provided is wrong, or they are right and i have no idea why.

Please help!

 

[pka]

Into how many distinct, equally sized (8 per group) groups can 32 unique cards be sparated, if 4 of those cards have to be together (eg. in the same group).

My solution was that first we make a set with 4 elements out of 28, because that is the set in which those 4 cards go, then we make a set with 8 elements out of 24, then a set with 8 elements out of 16 and then finally a set of 8 elements out of 8 elements (which is 1). We multiply these so we got something like: (28|4) * (24|8) * (16|8) * (8|8), now this would be correct if the order of the groups would matter, but it doesn't so we divide the whole thing with 4!, because there are 4 groups so we finally get: ((28|4) * (24|8) * (16|8) * (8|8)) / 4!

Now apparently my soulution is fine, except for the fact that we have to divide by 3!, according to the solution.
I'm either right and the solution provided is wrong, or they are right and i have no idea why.

Your notation is not standard. I don't know what (24|8) means.

In standard notation the answer is \(\displaystyle \dbinom{28}{4}\cdot\dfrac{24!}{(8!)^3\cdot3!}\)
This is a mixed partition question. In involves one ordered cell and three unordered cells.
That is where the 3! comes in.

 

[WinterGraschp]

Your notation is not standard. I don't know what (24|8) means.

In standard notation the answer is \(\displaystyle \dbinom{28}{4}\cdot\dfrac{24!}{(8!)^3\cdot3!}\)
This is a mixed partition question. In involves one ordered cell and three unordered cells.
That is where the 3! comes in.

I wasn't really sure how to add standard notation, and even after searching a bit, i came up empty handed, i apologize for the inconvinience, I'd also be grateful if you could point me somwhere where they explain how to use standard notation.

How do I know which cells are ordered and which are not? Do some of them count as ordered if they are the same size? And if some of the cells are ordered, then i have to exclude repetitions by diving as in this example by 3!, because there are 3 ordered cells?

 

[WinterGraschp]

So if I have a set with 14 elements and I would want to calculate the number of partitions whose underlying partition is [4,4,2,2,2], then would the solution look like:\(\displaystyle \dfrac{\dbinom{14}{4}\cdot\dbinom{10}{4} \cdot \dbinom{6}{2} \cdot\dbinom{4}{2}\cdot\dbinom{2}{2}}{2! \cdot 3!}\)

 

[pka]

So if I have a set with 14 elements and I would want to calculate the number of partitions whose underlying partition is [4,4,2,2,2], then would the solution look like:\(\displaystyle \dfrac{\dbinom{14}{4}\cdot\dbinom{10}{4} \cdot \dbinom{6}{2} \cdot\dbinom{4}{2}\cdot\dbinom{2}{2}}{2! \cdot 3!}\)

You have the correct idea. Now tell me why
\(\displaystyle \dfrac{\dbinom{14}{4}\cdot\dbinom{10}{4} \cdot \dbinom{6}{2} \cdot\dbinom{4}{2}\cdot\dbinom{2}{2}}{2! \cdot 3!}=\dfrac{14!}{(4!)^2(2!)^3(2!)(3!)}~?\)

 

[stapel]

I wasn't really sure how to add standard notation, and even after searching a bit, i came up empty handed....

Not a prob. Just be sure to define your notation when it's something that you've made up. That way, we're all on the same page. Thank you!