Answer:
(a) Number of sets B given that
- A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
- A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)
(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.
Step-by-step explanation:
<h3>(a)</h3>
Let 
denote the 20 elements of set X.
Let 
denote elements of set X that are also part of set A.
For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain .
For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from .
![\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bc%7Ccccccccc%7D%5Ctext%7BMembers%20of%20X%7D%20%26%20x_1%20%26%20x_2%20%26%20%5Ccdots%20%26%20x_%7B10%7D%20%26%20x_%7B11%7D%20%26%20%5Ccdots%20%26%20x_%7B20%7D%5C%5C%5B0.5em%5D%5Cdisplaystyle%5Ctext%7BMember%20of%7D%5Catop%5Cdisplaystyle%5Ctext%7BSet%20A%3F%7D%20%26%20%5Ctext%7BYes%7D%26%5Ctext%7BYes%7D%26%5Ccdots%20%26%5Ctext%7BYes%7D%26%20%5Ctext%7BNo%7D%20%26%20%5Ccdots%20%26%20%5Ctext%7BNo%7D%5C%5C%5B0.5em%5D%5Cdisplaystyle%5Ctext%7BMember%20of%7D%5Catop%5Cdisplaystyle%5Ctext%7BSet%20B%3F%7D%26%20%20%5Ctext%7BYes%7D%26%5Ctext%7BYes%7D%26%5Ccdots%20%26%5Ctext%7BYes%7D%26%20%5Ctext%7BMaybe%7D%20%26%20%5Ccdots%20%26%20%5Ctext%7BMaybe%7D%5Cend%7Barray%7D)
.
For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus 
possibilities for set B.
In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves 
possibilities.
<h3>(b)</h3>
Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.
Start by considering the case when set A and set B are indeed disjoint.
![\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bc%7Ccccccccc%7D%5Ctext%7BMembers%20of%20X%7D%20%26%20x_1%20%26%20x_2%20%26%20%5Ccdots%20%26%20x_%7B10%7D%20%26%20x_%7B11%7D%20%26%20%5Ccdots%20%26%20x_%7B20%7D%5C%5C%5B0.5em%5D%5Cdisplaystyle%5Ctext%7BMember%20of%7D%5Catop%5Cdisplaystyle%5Ctext%7BSet%20A%3F%7D%20%26%20%5Ctext%7BYes%7D%26%5Ctext%7BYes%7D%26%5Ccdots%20%26%5Ctext%7BYes%7D%26%20%5Ctext%7BNo%7D%20%26%20%5Ccdots%20%26%20%5Ctext%7BNo%7D%5C%5C%5B0.5em%5D%5Cdisplaystyle%5Ctext%7BMember%20of%7D%5Catop%5Cdisplaystyle%5Ctext%7BSet%20B%3F%7D%26%20%20%5Ctext%7BNo%7D%26%5Ctext%7BNo%7D%26%5Ccdots%20%26%5Ctext%7BNo%7D%26%20%5Ctext%7BMaybe%7D%20%26%20%5Ccdots%20%26%20%5Ctext%7BMaybe%7D%5Cend%7Barray%7D)
.
Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus possibilities for a set B that is disjoint with set A.
There are 20 elements in X so that's 
possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only 
possibilities left.
