vestal niche
In how many ways can 4 pieces of candy be given to 3 kids if the candy is distinguishable, and the kids are indistinguishable?
I'm understanding that there are four cases:
C1) {4, 0, 0}
C2) {3, 1, 0}
C3) {2, 2, 0}
C4) {2, 1, 1}
Case three stumps me
rose cradleBOT
In how many ways can 4 pieces of candy be given to 3 kids if the candy is distin...
+close
vestal niche
Help as soon as possible would be appreciated, but this is just a problem I stumbled upon and struggle to wrap my head around...
vestal niche
I would also assume that case 3 would equal 6, but the correct answer does not seem to be 17
@mild raptor you have any suggestions on this one?
vestal niche
Nvm solved it
dawn coyote
In how many ways can 4 pieces of candy be given to 3 kids if the candy is distin...
If the candy is distinguishable, that means we don't have four pieces of candy, we have candies A, B, C, and D.
Are we required to distribute all of the candies to the children, or do we have to do cases where we distribute, say, only A, only B, only C, only D, only A and B... etc?
sturdy girder
In how many ways can 4 pieces of candy be given to 3 kids if the candy is distin...
Case 3 will be 4!/(2!2!)
All your other cases are right, you have to add these cases to get the total number of ways
sturdy girder
Help as soon as possible would be appreciated, but this is just a problem I stum...
I find it cute that you called it rap
vestal niche
I find it cute that you called it rap
sturdy girder
<:flush:1111561099809132585>
Did you solve it?
sturdy girder
I think I did
What was your way to do it?
vestal niche
But could you explain a little more about why 4!/2!2!
sturdy girder
How is 4!/2!=6?
sturdy girder
That's what you wrote in case 4 of your answer
sturdy girder
Wait I'll tell you, you wrote 4!/2!=6
vestal niche
The repeats are because if I have candies AB
CD is forced
And as such
CD AB is forced
Right?
Am I going about this the wrong way
sturdy girder
I don't understand your explaination here, so nk idea
What I think is $1 + \frac{4!}{3!1!} + \frac{4!}{2!2!} + \frac{4!}{2!1!1!} = 1+4+6+12 = 23$
night jungleBOT
fäf kaka
vestal niche
Are we required to distribute *all* of the candies to the children, or do we hav...
Yes
vestal niche
What I think is $1 + \frac{4!}{3!1!} + \frac{4!}{2!2!} + \frac{4!}{2!1!1!} = 1+4...
I got 1+4+6+3…
sturdy girder
Yes
Well do you know the method where we find number of ways $r$ digits add up to $n$?
vestal niche
Unsure
sturdy girder
If we go by this, then we get ${4+3-1 \choose 4-1} = {6 \choose 3} = 20$
vestal niche
That doesn’t work in this case because
night jungleBOT
fäf kaka
vestal niche
The candies are distinguishable
But the children were not
If it were the other way around I believe it would work though
Atleast that’s what I remeber
sturdy girder
But the children were not
Wait, that's why it should work actually
vestal niche
Stars and bars work
When k is distinguishable
But it isn’t in this case
The candies are
It’s like saying put 0 kids in a candy
Or one kid
No?
I could just be wrong.
sturdy girder
**fäf kaka**
Although this is what I got by doing it by the way I always do
vestal niche
The correct answer does seem to be 14
@dawn coyote could you explain your process to me?
vestal niche
Yes
All candies must be distributed amongst the children but any child is not required to have candy
dawn coyote
<@352737152154468352> could you explain your process to me?
Well, it's not complete, but I was thinking about how we would construct the set of solutions.
So, like, let's define the set S = {A, B, C, D}.
vestal niche
Yes
dawn coyote
Then we want to construct the set X of all sets Y such that:
vestal niche
Why would the absolute value of y be 3?
dawn coyote
Three children, right?
dawn coyote
...wait. Okay, that doesn't work, because Y = {{A, B, C, D}, 0, 0} should be in X, but isn't, because by extensionality it actually only has a magnitude of 2.
vestal niche
dawn coyote
Well, whatever. We can construct X as I've defined it and just add one to the cardinality, which is the part of X we care about.
vestal niche
Alright, I think I kinda get it so far
I’m thankful for your hope ur unfortunately I must go eat dinner, um hopefully you could explain your solution whenever you feel as if you had time and I can read it when I get back?
sturdy girder
I’m thankful for your hope ur unfortunately I must go eat dinner, um hopefully y...
You can just make all cases and see how many you get
dawn coyote
Hmm. Okay, so the way I think about it is, there's one quartet in P(S), right? Which corresponds to one child getting all four candies.
There's four triplets.
sturdy girder
Hmm. Okay, so the way I think about it is, there's one quartet in `P(S)`, right?...
Whatever you do, just make sure you get 23
dawn coyote
Corresponding to one child getting three of the candies and a second child getting the fourth.
Since we don't care which child gets which set of candies, there is only one distribution for each triplet.
Then there's six couples. This is where it gets complicated, because each couple can be paired with either a couple or two singles.
So, let's see. That's six couple/single/single sets and three couple/couple pairs (since couple pairs are isomorphic on permutation, thus the six couple pairs get divided in half). So that's 1 + 4 + 6 + 3 = 14. And we can't do anything with the singletons because every possible combination of singletons was already accounted for.
vestal niche
the difference is that the kids we are distributing to are indishtinguable
so no
how what
you cant put
a kid in a candy
which is also why stars and bars dont work
dawn coyote
...what?
...to... candies...?
But each candy can only be distributed to one kid.
...no, I'm saying 3000 doesn't work. That's distributing Candy A to all three kids, and candies B, C, and D to none, right?
But Candy A can only go to one kid, and B, C, and D must go to a kid.
I mean, that's what I did. I'm sure there's probably a more efficient way, but it would have to come out to the same answer.
And your approach is irreparably flawed.
No offense.
...it's literally the last thing said in the channel before you got here.
dawn coyote
What "211 case"?
dawn coyote
That doesn't help me.
That would be a doubleton and two singletons. There are six doubletons (4C2), and each has only a single pair of singletons that complete the distribution, so... there's six.
...what?
No.
Where's the other six coming from?
There are only three of those.
dawn coyote
...because {{A, B}, {C, D}} = {{C, D}, {A, B}}
dawn coyote
...why 4c2?
dawn coyote
...but we were talking about 2, 2, 0.
Except, again, {{A, B}, {C, D}, 0} = {{C, D}, {A, B}, 0}.
dawn coyote
^
vestal niche
As long as any two candies are chosen first, it forced to the other person having two candies, as such they have the same probability and why divide by two
In my mind that’s how I saw it
dawn coyote
As long as any two candies are chosen first, it forced to the other person havin...
Because the case where Child 1 gets A, B and Child 2 gets C, D is identical to the case where Child 1 gets C, D and Child 2 gets A, B. {{A, B}, {C, D}, 0} = {{C, D}, {A, B}, 0}.
vestal niche
***Because the case where Child 1 gets A, B and Child 2 gets C, D is identical t...
Is it not basically the same thing
dawn coyote
Is it not basically the same thing
Yes. It is the same thing. Which is why we only count it ONCE, NOT TWICE, LIKE WHAT YOU'RE DOING.
dawn coyote
Wait.
vestal niche
?
dawn coyote
Sorry. Didn't realize I was explaining to two different people.
