How many combinations can be made if you have 6 items, and each of these item can have a value of 1-100?

Question

So i have 6 different flavours.

I intend to mix these flavours together where the total amount would be 100 % when all the 6 different flavours has been counted.

This means for example that one combination can look like this flavour 1 can have 20 % in there, flavour 2 = 40 %, flavour 3 = 15 %, flavour 4 = 15 %, flavour 5 = 6 %, flavour 6 = 4 %. That setup would together make 100 % and 1 combination has been made.

How do i calculate how many possible combinations you can make with this? I am only interested in how to calculate "full numbers" so no 4,5 % for instance.

Answer 1

The answer depends on whether you require each of the six flavours to be present (that is, at least 1%).

Assuming that you do indeed intend each flavour to be present: Consider 100 little cups waiting to take their flavour portion, representing 1% each. Now insert marks between the cups to show where to change to the next flavour - you need 5 such marks, to insert into 99 different positions.

Thus the answer is $\binom {99}{5} = \frac{99!}{94!\cdot5!}$.

For the other case where some flavours can be absent, this means you can have multiple marks in one space. Now the result looks like arrange 100 cups and 5 marks, which means we have 105 places and we choose 5 of those places to be flavour-change marks.

For this case the answer is $\binom {105}{5} = \frac{105!}{100!\cdot5!}$.

Answer 2

In case you need a proof for the binomial stated in the comment above or in the answer above you may consider the following sketch:

Suppose you have $100$ spoons in a row represented here by the letter "o", namely something like this ooooooooooooooooooooooooooooooooooooooo........ooo then in how many ways can we separate this row with 5 vertical bars "\" (the number of "o" between the bars represent the amount of each flavour)? One example would be: ooooo\oooooo\oooo\ooooo\ooooo\oooooooo...o For sure we have $99$ positions if we do not want a bar to be placed outside of the spoon row. That is why we have the binomial $\binom {99}{5} = \frac{99!}{94!\cdot5!}$.