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.

can you tell the number of positive integer solutions of the equation $a+b+c+d+e+f=100$, ? — dmtri, Sep 05 '18 at 15:42
I dont understand 100 %. The outcome will always be 100%. So it cannot be less than 100 % in total. — Carlos Rodrigez, Sep 05 '18 at 15:54
So for instance another combination is, a = 100 %, b = 0 %, b = 0%, c = 0%, d = 0%, e = 0%, f = 0% — Carlos Rodrigez, Sep 05 '18 at 15:55
ok you may find then the number of non negative integer solutions — dmtri, Sep 05 '18 at 15:57
I can do 99 choose 5 on here http://www.wolframalpha.com/input/?i=99+and+6+choose+5 but how do i setup like u did? — Carlos Rodrigez, Sep 05 '18 at 18:50

Joffan · Accepted Answer · 2018-09-05T16:48:26.557

2

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 $\binom {105}{5} = \frac{105!}{100!\cdot5!}$.

edited Sep 05 '18 at 16:48

answered Sep 05 '18 at 15:54

Joffan

39,627

They do not have to be present no, but a quantity of 100% always has to be filled – Carlos Rodrigez Sep 05 '18 at 15:58
So for instance another combination is, a = 100 %, b = 0 %, b = 0%, c = 0%, d = 0%, e = 0%, f = 0% – Carlos Rodrigez Sep 05 '18 at 15:58
Hi Carlos, there is an inconsistency here because the question says "when all 6 different flavors are in" – Phil H Sep 05 '18 at 16:05
@CarlosRodrigez extended answer to cover the cases you describe. – Joffan Sep 05 '18 at 16:49
@PhilH Ah my mistake, updated that now – Carlos Rodrigez Sep 05 '18 at 17:05

score 0 · Answer 2 · answered Sep 05 '18 at 18:12

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 $\binom {99}{5} = \frac{99!}{94!\cdot5!}$.

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

2 Answers2