How Many Ways to Build a 6-Pack There is a beverage company here that claims to have a selection of 200 different beers. They have a special deal where you can build your own six pack at a discount. They advertise that there are 1.4B ways to build said six pack, and I am trying to determine if they're correct.
I thought that this would be a combination with repetition problem. You have $n=200$ items with $r=6$ choices. The formula $$x=(n+r-1)!/(r!(n-1)!)$$ yields $x=95,746,959,700$.
Then I tried to work backwards using $x=1400000000$. I plugged $$1400000000=n!(n-1)!(n-2)!(n-3)!(n-4)!(n-5)!$$ into Wolfram Alpha to get the solution. It showed that the alternate form is a gamma function $$1400000000 = Γ(x-4)Γ(x-3)Γ(x-2)Γ(x-1)Γ(x)Γ(x+1)$$ That solution is approximately $n=6.496$. My experience with combinatorics is limited to pre-calculus permutations and combinations, so gamma functions is somethings that I stumbled upon literally half an hour ago. So I ask: How many ways can you build a six pack with 200 beers?
Note: I understand that you could use $x=n^r$, because technically three Bud Lights and three Coors Lights are different from three Coors Lights and three Bud Lights. The answer for that permutation is $x=64,000,0,000,000$.
 A: Let $(a_1, \dots, a_l) \in \mathbb{N}^l$ be notation for $a_j$ bottles of the $j^{\text{th}}$ beer.
We need to consider how many different types of beer there are in the six pack. Obviously, there could be one, two, three, four, five, or six. Let's consider each case individually.
One type: The only way this can occur is if there are six bottles of the same beer. There are ${200 \choose 1} = 200$ ways of choosing that type.
Two types: There are three ways this can occur: $(5,1)$, $(4,2)$, and $(3,3)$ - note, the order of the beers (i.e. which type you call first and which you call second) doesn't count. There are ${200 \choose 2}$ ways of choosing the two types of beer. There are three possibilities for the six pack that can be made with those two types. So there are ${200 \choose 2}\times 3 = 59,700$ ways of creating a six pack with two types of beer.
Three types: There are two ways this can occur: $(4,1,1)$ and $(3,2,1)$. There are ${200 \choose 3}$ ways of choosing the three types of beer. There are two possibilities for the six pack that can be made with those three types. So there are ${200 \choose 3}\times 2 = 2,626,800$ ways of creating a six pack with three types of beer.
Four types: There are two ways this can occur: $(3,1,1,1)$ and $(2,2,1,1)$. By a similar argument, there are ${200 \choose 4}\times 2 = 129,369,900$ ways of creating a six pack with four types of beers.
Five types: There is only one way this can occur: $(2,1,1,1,1)$. There are ${200 \choose 5} = 2,535,650,040$ ways of creating a six pack with five types of beers.
Six types: There is only one way this can occur: $(1,1,1,1,1,1)$. There are ${200 \choose 6} = 82,408,626,300$ ways of creating a six pack with six types of beers.
Therefore the number of possible six packs that can be made is $${200 \choose 1} + {200 \choose 2}\times 3 + {200 \choose 3}\times 2 + {200 \choose 4}\times 2 + {200 \choose 5} + {200 \choose 6} = 85,076,332,940.$$
So, in short, they are not correct.

More generally, suppose you have $n$ types of object and you wish to choose $k$ objects which can include any number of each type. How many ways are there of doing this? By the same logic, there are $$\sum_{i=1}^k {n \choose i}\times P(k, i)$$ ways where $P(k, i)$ is the number of partitions of $k$ into $i$ parts, i.e. the number of ways $k$ can be written as a sum of $i$ positive integers.
A: I distrust advertising numbers like this.  The original Rubik's cube promised billions of positions.  They were technically correct, the correct number is $43,252,003,274,489,856,000$ or 43 billions of billions.  To get it exactly right, you would have to make a list of all the partitions of $6$, figure the multinomial coefficient for each one, and add them up.  But calculation of ${200 \choose 6}=82,408,626,300$, the number with six distinct beers, already shows there are many more than $1.4E9$.
A: If we count permutations as distinct combinations, then it should be $\frac{200!}{194!}$ which is in the order of $10^{13}$. Otherwise it should be $\binom{200}{6}$, which is in the order of $10^{10}$.
