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You want to put 10 cokes into a cooler. There are 6 brands of coke. Each brand has more than 10 cans. How many configurations are there if positions do not matter.

This is just grim

N. F. Taussig
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Trajan
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  • "There are 6 brands of coke". No, there aren't! "Coke" is a trade-mark of the Coca-Cola company. There may have been 6 brands of cola but there is only one brand of "Coke". (Oops, wait! Since "coke" has a small c, do you mean "6 brands of cocaine?) – user247327 Nov 10 '20 at 15:44
  • This is a combinations with repetition problem since you are selecting $10$ objects from $6$ types of objects. – N. F. Taussig Nov 10 '20 at 15:46
  • @user247327 it is used as a "genericized trademark" Here in Atlanta, everything is referred to as "coke" whether produced by the CocaCola company or otherwise. We say "coke" like people in the midwest say "pop." – JMoravitz Nov 10 '20 at 15:50
  • As for the math of this, as N.F.Taussig already alludes, this is a routine textbook example of using stars-and-bars. – JMoravitz Nov 10 '20 at 15:52
  • @JMoravitz its not clear to me how its stars and bars# – Trajan Nov 10 '20 at 16:01
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    @Trajan let $x_1$ be the number of drinks you use of the "first brand", let $x_2$ be the number of drinks of the "second brand", etc... Your asking how many different ways you can fill the cooler is equivalent to asking how many tuples $(x_1,x_2,\dots,x_6)$ there are with $x_i\geq 0$ for each $i$ and $x_1+x_2+\dots+x_6=10$ – JMoravitz Nov 10 '20 at 16:02
  • @JMoravitz i still dont get it because in this problem we have 10 cans of each brand? – Trajan Nov 10 '20 at 16:08
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    Note closely how the phrase "each drink has more than 10 cans available" affects the problem. Since the number of available cans of each drink is more than the total number of drinks we will use, we don't need to worry about using too many of any type of drink. If we were told instead that we were limited to using at most five of each type of drink that would have been a more complicated problem (by a bit) since among those outcomes counted by stars-and-bars some would have been impossible as we used too many of certain flavors. – JMoravitz Nov 10 '20 at 16:09
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    All that the phrase "each drink has more than 10 cans available" is meant to signify here is that we don't have to care about how many cans we have available for each drink since it will always be enough... no outcome counted in stars-and-bars will be impossible due to having used too many of a particular drink. – JMoravitz Nov 10 '20 at 16:11
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    Translating it to the diophantine equation representation of the problem, letting $A_1$ be the number of available cans for the drink of the first type, $A_2$ the number of available cans of the drink of the second type, etc... since we are limited in the number of resources of each type available we really have $\begin{cases}x_1+x_2+\dots+x_6=10\0\leq x_1\leq A_1\ 0\leq x_2\leq A_2\\vdots\0\leq x_6\leq A_6\end{cases}$... but since we are told $A_i>10$ for each $i$ we recognize this as being able to be simplified to $\begin{cases}x_1+x_2+\dots+x_6=10\0\leq x_i\end{cases}$ – JMoravitz Nov 10 '20 at 16:14
  • @JMoravitz i get you but i cant exactly see the justification right now – Trajan Nov 10 '20 at 16:22
  • I don't understand what there isn't to get... you don't see the justification for what exactly? Why the two diophantine equations are equivalent? Why counting the number of solutions of a diophantine equation like this is equivalent to counting the number of possible ways of filling the cooler? Why the solution to the counting problem of counting number of integer solutions to the diophantine equation is given by stars and bars? – JMoravitz Nov 10 '20 at 16:25
  • Let us continue this discussion in chat. – JMoravitz Nov 10 '20 at 16:26

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