Question

How many elements are in the union of four sets if each of the sets has 96 elements, each pair of sets share 52 elements, each triple of sets shares 22 elements and there are 6 elements in all four sets.

Answer 1

154 elements

Explanation:

Let the four sets be X1, X2, X3, X4.

Each of the four sets has 96 elements

|X1| = 96

|X2| = 96

|X3| = 96

|X4| = 96

They can be paired in the following ways

|X1 n X2| , |X1 n X3|, |X1 n X4|, |X2 n X3|, |X2 n X4|, |X3 n X4|

Each pair of sets contains 52 elements each

|X1 n X2| = 52

|X1 n X3| = 52

|X1 n X4| = 52

|X2 n X3| = 52

|X2 n X4| = 52

|X3 n X4| = 52

The triple sets are |X1 n X2 n X3|, |X1 n X2 n X4|, |X1 n X3 n X4|,|X2 n X3 n X4|

Each triple set has 22 elements

|X1 n X2 n X3|=22

|X1 n X2 n X4|= 22

|X1 n X3 n X4|=22

|X2 n X3 n X4|=22

All the four sets have 6 elements

|X1 n X2 n X3 n X4| = 6

Using the principle of inclusion,

|X1 υ X2 υ X3 υ X4|= |X1|+ |X2|+ |X3|+ |X4|- |X1 n X2| - |X1 n X3| - |X1 n X4| -|X2 n X3| - |X2 n X4| - |X3 n X4|+ |X1 n X2 n X3|+ |X1 n X2 n X4|+|X1 n X3 n X4|+|X2 n X3 n X4- |X1 n X2 n X3 n X4|

= 96 + 96 + 96 + 96 – 52 – 52 – 52 – 52 – 52 – 52 + 22 + 22 + 22 + 22 – 6

= 4(96) -6(52) +4(22) – 6

= 384 – 312 + 88 – 6

= 154