A number of applications of the algebra of sets, particularly in probability theory, depend upon the number of elements in a set. We will denote the number of elements in a set X by n(X).

Suppose that for two sets A and B we know that n(A) = 50 and n(B) = 100. What can be said concerning n(A + B) and n(AB)? It is apparent that if A and B have no elements in common, n(A + B) = 150 and n(AB) = 0. In the general case, we can say only that 100 ≦ n(A + B) ≦150, where n(A + B) = 100 if and only if A ⊆B. Like- wise, 0≦n(AB) ≦ 50, where n(AB) = 50 only in the case A ⊆ B.

In general, if X and Y have no elements in common, we say that X and Y are disjoint sets and the formula n(A + B) = n(A) + n(B) holds. In all cases the following theorem holds.

THEOREM 1. If X and Y are any two sets, then

                                n(X + Y) = n(X) + n(Y) - n(XY).

Proof. Since XY and XY' are disjoint sets and X = XY+ XY', it follows that n(X) = n(XY) + n(XY'). Similarly, n(Y) = n(XY) + n(X'Y). Adding these two equations, we obtain n(X) +n(Y) = n(XY')  + n(X'Y) + 2n(XY), or n(XY') + n(X'Y) = n(X) +n(Y) - 2n(XY).

Next we note that XY', X'Y, and XY are disjoint sets, and

X+Y=X(Y+Y')+Y(X+X') =XY+XY'-' XY+X'Y = XY+XY'+X'Y.

Hence n(X + Y) = n(XY) + n(XY') + n(X'Y) and, substituting from above, n(X + Y) = n(X) + n(Y) - n(XY), which completes the proof.

COROLLARY.

n(X + Y + Z) = n(X) + n(Y) + n(Z) - n(XY) - n(XZ) - n(YZ) + n(XYZ)

for any three sets X, Y, and Z.

The results of the theorem and corollary could be extended to include the case of four or more sets, but the resulting formulas become increase ingly unwieldy. As an exercise, the student should attempt to write the general formula which holds for m sets. This formula is not often used,  although the generalization of the method used in the first line of the proof of Theorem 1 is of considerable use in probability, as we shall see. This generalization is given in the following theorem:

THEOREM 2. If Y₁, Y₂, ... , Y_m are arbitrary sets which are mutually

disjoint and have the property that Y₁ + Y₂ + ……..+ Y_m = 1, then for any set X, n(X) = n(XY₁) + n(XY₂) + …….+ n(XY_m).

Proof. X=X(1)=X(Y_I+Y₂+...+Y_m) =XY₁+XY₂+...+ X Y_m where the sets XY₁, XY₂, ... , X Y_m are mutually disjoint.

From this the theorem follows.

ExAMPLE 1. (Taken from the Joint Associateship Examination for Actuaries,  1935, Part 5, question 9B.) Certain data obtained from a study of a group of 1000 employees in a cotton mill, as to their race, sex, and marital status, were unofficially reported as follows: 525 colored lives; 312 male lives; 470 married lives; 42 colored males; 147 married colored; 86 married males; 25 married colored males. Test this classification to determine whether the numbers reported in the various groups are consistent.

Solution. Let C be colored lives, Al be male lives, M' be married lives. Then

n(C + M + W) = n(C) + n(M) + n(W) - n(CM) - n(MW) - n(CW) + n(CMW)

                         = 525 + 312 + 470 - 42 - 86 - 147 + 25

                              = 1057.

The conclusion is that the data are inconsistent, since the data referred to only 1000 employees.

Of course, in this example, it is possible that this check might have given a number less than 1000 even though the data were inconsistent.

(See Problem 3 below.) In such a case, and in fact in any problem where the number of elements in two or more sets and their intersections are of interest, it is helpful to draw an appropriate Venn diagram and fill in the number of elements in each of the disjoint sets represented in the diagram.

EXAMPLE 2. The following information is given concerning the number of elements in the subsets A, B, C of a certain set with 200 elements: n(A) = 70,

n(B) = 120, n(C) = 90, n(AB) = 50, n(AC) = 30, n(BC) = 40, and n(ABC)  = 20. Find (a) n(A + B), (b) n(A + B + C), (c) n(A'BC), and (d) n(AB'C').

FiG. 1-1. Numbered Venn diagrams for Example 2.

Solution. The appropriate Venn diagram is given in Fig. 1-1. First, each of the regions in the diagram is labeled appropriately. Next, beginning with region

ABC, the correct number of elements is filled in. Here the number is given as 20.

Now the fact that n(ABC') = 30 is determined by subtracting n(ABC) from n(AB), since n(AB) = n(ABC) + n(ABC'). By continuing in this way, the number in each region is easily determined. If the data are inconsistent, the computed number of elements in some region will be negative, indicating that the given figures cannot be correct. If no inconsistency is present, any problem related to the diagram can be solved by inspection. The answers required in this example are: (a) 140, (b) 180, (c) 20, and (d) 10.