MHB How Many Unique Chocolate Combinations Can You Create?

[Hypatia1]

There are 10 different chocolates and you want to buy three of them. However, you cannot pick a pair of chocolate more than once. How many different choice you can make??

 

[Evgeny.Makarov]

By "chocolates" do you mean 10 candy types or 10 individual candies?

 

[Hypatia1]

Ten differrent type of chocolates
Lets say ten different brands

 

[Evgeny.Makarov]

you want to buy three of them

And what about this phrase? Can I buy, say, 7 chocolates of 3 different types or does it have to be 3 individual chocolates?

you cannot pick a pair of chocolate more than once.

This is also unclear. If I pick a pair twice, I have 4 chocolates. Why is this a restriction if I need to buy just 3? Or, if I have to buy any number of chocolates of 3 different types, can I buy 3 chocolates of type 1? I have 1 pair and another single chocolate, so no pair is picked twice.

Please describe the problem statement as clearly as you can.

 

[Hypatia1]

So let us say that we have the following chocolate brands
A B C D E
The number of three groups that can be formed with these brands will be 10. The order of your selection does not important and all three chocolate brands in each group should be different as the following
ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE
and let us say you buy A B C. From that point, but you can not buy chocolote groups involving A B, A C, or B C because you have chosen those pair of brands when you selected the ABC group. You cannot also buy A A A or A A B. Because that is not an option, as you see there will be 10 groups as listed above. When you select ABC, you can only buy ADE from the above list.

 

[Kansas Boy]

I'm not sure exactly what you are saying! It is true that, if we cannot have the same letter twice then we can choose any of the five letters first, then any of the remaining four, the any of the remaining 3 so 5(4)(3)= 60 different ways. But that includes the same three letters in different orders, such as ABC and BAC which apparently you do not want. There are 3!= 3(2)(1)= 6 different orders in which you can order three different ways. To discount those, divide by 3!= 6 to get 60/6= 10.

Now, what is the difference between "selecting" ABC and "buying" ADE?

 

[Hypatia1]

Selecting and buying mean the same. I used them interchangeably. Do you familiar with the social golfer problem? It is a very famous combinatorial optimization problem. Maybe I could not express as clearly as possible. But actually what I am asking is very similar to that of the social golfer problem. You can check it from the link below. https://en.m.wikipedia.org/wiki/Social_golfer_problem

But anyway, I want to write this problem here:

A group of 32 golfers plays golf once a week in groups of 4. Schedule these golfers to play for as many weeks as possible without any two golfers playing in the same group more than once.

But this problem, as you may notice, involves additional variable:week. Which I am not interested in. So let me phrase the problem again.

A group of 16 players plays golf once a week in groups of 4. Find the maximum number of groups for these golfers to play without any two golfers playing in the same group more than once.

How can we solve it?