If there are 100 different flavors, how many possible mixtures can be created?

Question

If I have 100 different flavors, how many possible different flavors can be created?

Answer 1

This is a question about an object called the Power Set.  In set theory we construct an object called the Power Set of a given set S, dented P(S), by considering the set of ALL subsets of S.  In this example, lets consider the set S to consist of our 100 flavor options (essentially ingredients for the mixtures).  

 

Each possible mixture we can create is a subset of S, namely we choose some number of specific flavors to mix together.  The beauty of creating this Power Set, P(S) is we can easily count how many subsets of a set there are.  This is a well known (and easily proven result) that P(S) has 2^N subsets, where N is the number of elements of S.  One way to easily see this is that each element in S can either be in a subset or not in a subset, this contributes a factor of 2 to the number of total subsets, for each of the elements of S...and since the choice of inclusion for a given element is independent from your choices for the other elements, multiplicativity gives us the 2^N is the number of combinations of N elements we can create.  

 

We need to be a little careful when we count the flavors we can create, because the above object P(S) includes the trivial, or empty Set...namely it counts the possibility of creating a mixture consisting of NO flavors at all...this technically shouldn't count for our problem, and so the answer we are looking for is 2^100-1.

 

 

This is just one way to combinatorially count the flavor combinations.  There are other techniques involving summing over the combinations that include a specific number of flavors, but I believe this is a more straightforward technique for this problem.

 

I hope this helps.

 

Answer 2

One more comment...

 

The question said, "how many possible mixtures can be created?"  And "how many possible different flavors can be created?"

 

As I see it, the flavors that consist of any one of the original 100 flavors alone are not "mixtures" and they are not "created."  I don't think I'm being overly literal to interpret the question this way -- it seems to be a reasonable interpretation of the two ways the question was asked.

 

Therefore, you have to subtract 100 from the number of non-null power sets:  2¹⁰⁰ - 101

 

 

Answer 3

Hi Jordan,

 

Short answer - very many!

 

The formula for the number of different combinations of K items drawn N items is C(K,N) = N! / [ (N-K)! x K!] where "!" is the factorial operator, which for any integer N>0, N! is defined to be N x (N-1) x (N-2) x ..... x 1 and 0! is defined to be 1.

 

Here we want the number  of combinations of all 100 flavors plus the number of different combinations that use 99 of the 100 flavors plus the number of different combinations of 98 of the 100 flavors and so on down to the number of different combinations of 1 flavor of the 100 different flavors. In the formula, we want to hold the N constant at 100, compute the formula value for each K running from 100 to 1, and add these 100 different results.

 

This is definitely a problem for Excel. In Excel, the COMBIN(number, number_chosen) function gives the number of different combinations from number items (here 100 flavors) chosen number_chosen at a time (here we want number_chosen to vary from 100 to 1.

 

So in column A of a spreadsheet, have the numbers running from 100 to 1. In column B of the spreadsheet in the first row, use the formula "=COMBIN(100,A1)" - then drag (i.e. copy) the formula down column B. After completing that operation, column B will contain the answers for the number of combinations from 100 flavors of column A number of flavors taken at a time. You can then total the number of flavors. You will obtain the following two columns and total.

A B

100 1
99 100
98 4950
97 161700
96 3921225
95 75287520
94 1192052400
93 16007560800
92 1.86088E+11
91 1.90223E+12
90 1.73103E+13
89 1.4163E+14
88 1.05042E+15
87 7.11054E+15
86 4.41869E+16
85 2.53338E+17
84 1.34586E+18
83 6.65013E+18
82 3.06645E+19
81 1.32342E+20
80 5.35983E+20
79 2.04184E+21
78 7.33207E+21
77 2.48653E+22
76 7.97761E+22
75 2.42519E+23
74 6.99575E+23
73 1.91735E+24
72 4.99881E+24
71 1.24108E+25
70 2.93723E+25
69 6.63246E+25
68 1.43013E+26
67 2.94692E+26
66 5.80717E+26
65 1.09507E+27
64 1.9772E+27
63 3.42003E+27
62 5.67005E+27
61 9.01392E+27
60 1.37462E+28
59 2.01164E+28
58 2.82588E+28
57 3.81165E+28
56 4.93782E+28
55 6.14485E+28
54 7.3471E+28
53 8.44135E+28
52 9.32066E+28
51 9.89131E+28
50 1.00891E+29
49 9.89131E+28
48 9.32066E+28
47 8.44135E+28
46 7.3471E+28
45 6.14485E+28
44 4.93782E+28
43 3.81165E+28
42 2.82588E+28
41 2.01164E+28
40 1.37462E+28
39 9.01392E+27
38 5.67005E+27
37 3.42003E+27
36 1.9772E+27
35 1.09507E+27
34 5.80717E+26
33 2.94692E+26
32 1.43013E+26
31 6.63246E+25
30 2.93723E+25
29 1.24108E+25
28 4.99881E+24
27 1.91735E+24
26 6.99575E+23
25 2.42519E+23
24 7.97761E+22
23 2.48653E+22
22 7.33207E+21
21 2.04184E+21
20 5.35983E+20
19 1.32342E+20
18 3.06645E+19
17 6.65013E+18
16 1.34586E+18
15 2.53338E+17
14 4.41869E+16
13 7.11054E+15
12 1.05042E+15
11 1.4163E+14
10 1.73103E+13
9 1.90223E+12
8 1.86088E+11
7 16007560800
6 1192052400
5 75287520
4 3921225
3 161700
2 4950
1 100
Total number of flavors 1.26765E+30

 

Here the "E+XX" is exponential notation, which means multiply the number that precedes the E by 10 raised to the XX power, or equivalently, shift the decimal point XX places to the right. This means that there are a total of

 

1.26765 x 10³⁰ different flavors that can be made.

 

Clearly a number this large needs some perspective. The number 10¹⁰⁰ is one googol. If Google had a googol of employees, then we would not be able to serve a unique flavor to each employee. However, the surface area of the earth (including water) is about 5.10072E+20 square millimeters, so if we got sloppy with an order for one cone for each flavor and dripped one square millimeter of ice cream on the ground for each cone, we could more than cover the earth with this many flavors.

 

Hope this helps. Feel free to ask questions.

 

Kind regards,

Hugh