what we need to do is find the cardinality of the following 8 sets:

1.A - (BUC) the people who watched "only A"

2.B - (AUC) (only B)

3.C - (AUB) (only C)

4.A∩B - C, the people who watched A and B, but not C.

5.A∩C - B (A and C, but not B)

6.B∩C - A (B and C, but not A)

7.A∩B∩C, people who watched all 3 shows

8.T - AUBUC, people who didn't watch ANY show (where T is our "total set").

we are given the following:

|A| = 55, |B| = 53, |C| = 55, |A∩B| = 35, |A∩C| = 28, |B∩C| = 23, |T| = 100, |T - AUBUC| = 10.

so we get subset 8, as a "freebie", there are 10 people who didn't watch any of the 3 shows.

this means that |AUBUC| = 90, 90 people watched at least one of the three shows.

now |AUBUC| = |A| + |B| + |C| - |A∩B| - |A∩C| - |B∩C| + |A∩B∩C|. we know all of these numbers except |A∩B∩C|, so we can calculate it:

90 = 55 + 53 + 55 - 35 - 28 - 23 + |A∩B∩C|

90 = 163 - 86 + |A∩B∩C|

90 = 77 + |A∩B∩C|

13 = |A∩B∩C|, which is subset 7.

now |A∩B| = |A∩B - C| + |A∩B∩C|, and again, we know 2 of these numbers, so we can solve for the 3rd:

35 = |A∩B - C| + 13

22 = |A∩B - C|, which is subset 4. similarly,

|A∩C| = |A∩C - B| + |A∩B∩C| ---> |A∩C - B| = 15, which is subset 5, and

|B∩C| = |B∩C - A| + |A∩B∩C| ---> |B∩C - A| = 10, which is subset 6.

finally |A| = |A - BUC| + |A∩B - C| + |A∩C - B| + |A∩B∩C|, so

55 = |A - BUC| + 22 + 15 + 13 ---> |A - BUC| = 5, which is subset 1. in a similar fashion:

|B| = |B - AUC| + |A∩B - C| + |B∩C - A| + |A∩B∩C| ---> |B - AUC| = 8, which is subset 2,

|C| = |C - AUB| + |A∩C - B| + |B∩C - A| + |A∩B∩C| ---> |C - AUB| = 17, which is subset 3.

this should enable you to finish the remainder of the equation.