I thought of the following problem. Suppose medical patients can have the following characteristics:
- Smoking Status: Often, Never, Sometimes
- Weight: Overweight, Underweight, Healthy
- Age: Child, Teenager, Adult, Senior
- Gender: Male, Female
- Blood Type: A, B, O, AB
Suppose you have a hospital that wants to make 3 clinics for these patients - for example, a potential group of clinics could be:
Group 1 (Example of Clinic-Patient Allocation - I just quickly made this allocation up, it might not be consistent with the "constraints" below):
Clinic 1: Smoking = (Often, Never), Weight = (Healthy), Age = (Adult, Child) , Gender = (Male), Blood Type = (A, B, O)
Clinic 2: Smoking = (Sometimes), Weight = (Overweight, Underweight), Age = (Senior), Gender = (Male, Female), Blood Type = (A, AB, B)
Clinic 3: Smoking = (Never), Weight = (Healthy), Age = (Child, Teenager), Gender = (Male), Blood Type = (A, AB)
Seeing as this is a combinatorial problem, there will likely exist "n" such groups - in which "n" is a very large number.
I am trying to determine how we can calculate the following number:
With these medical characteristics - how many "groupings" of clinics can be made such that the following constraints are met:
Constraint 1: A patient can ONLY be assigned to one of these clinics (e.g. a patient can not be assigned to both Clinic 1 and Clinic 2)
Constraint 2: No clinic is left empty (e.g. not allowed that Clinic 3 sees no patients)
Constraint 3: No patient can be "left behind" (e.g. If I am an Overweight Female Senior who Regularly Smokes and has Blood Type O - I must be able to know exactly which clinic is intended for me.)
In the end, if I am a Male Adult with Blood Type A who Never Smoked and Overweight - looking at these clinic definitions, I will know which clinic is meant for me.
- Is there some combinatorial formula that can be used to calculate the number of possible clinic groupings that can be made for 3 clinics with these medical conditions, such that the constraints are respected?
Thank you!
- Note: In this question, "comma" (,) means OR. For example:
Clinic 1 : Smoking = (Often, Never), Weight = (Healthy), Age = (Adult, Child) , Gender = (Male), Blood Type = (A, B, O)
Is the same as
Clinic 1 only accepts patients that : Smoking = (Often OR Never), Weight = (Healthy), Age = (Adult OR Child) , Gender = (Male), Blood Type = (A OR B OR O)
Of the $5$ listed characteristics that are being measured, the number of different properties associated with each characteristic is $3,3,4,2,4$ respectively.
So, you have (in effect) $3 \times 3 \times 4 \times 2 \times 4 = 288$ different kinds of patients.
You want to partition these patients into $3$ groups such that:
For simplicity, I will assume that of the $288$ possible different types of patients, each of these types has at least one patient in it.
If you were allowed to have one or two of the $3$ groups empty, then the computation would be
$$3^{(288)}, \tag1 $$
since there are $3$ choices for each of the $288$ different kinds of patients. To get the desired computation, you have to use (1) above as a starting point, and then subtract all of the distributions that violate the constraint that none of the groups can be empty.
This problem then takes on the properties of Inclusion-Exclusion, which is discussed here.
After reading that article, I suggest that you consult this answer for an explanation of and justification for the Inclusion-Exclusion formula.
Following the blueprint in the above link, for a set $E$ with a finite number of elements, I will let $|E|$ denote the number of elements in the set $E$.
Let $S$ denote the set of all possible distributions of the $288$ types.
Let $S_k$ denote the subset of $S$ that has group $k$ empty, where $k \in \{1,2,3\}.$
Let $T_0$ denote $|S|$.
Then, you want $T_0 - |S_1 \cup S_2 \cup S_3|.$
Let $T_1$ denote $|S_1| + |S_2| + |S_3|$.
Let $T_2$ denote $|S_1 \cap S_2| + |S_1 \cap S_3| + |S_2 \cap S_3|$.
Let $T_3$ denote $|S_1 \cap S_2 \cap S_3|.$
Then, the desired computation will be
$$T_0 - T_1 + T_2 - T_3.$$
Clearly,
$$T_0 = 3^{(288)}.$$
$$T_1 = \binom{3}{1} \times 2^{(288)}.$$
$$T_2 = \binom{3}{2} \times 1^{(288)}.$$
$$T_3 = 0.$$
Therefore, the final computation is :
$$ 3^{(288)} - \left[3 \times 2^{(288)}\right] + \left[3 \times 1\right].$$