Here is part (b):
The number of ways to choose three subsets A, B, C of S is 2^n * 2^n * 2^n = 2^(3n).
To find the number of ways that A is a subset of B, and B is a subset of C, we can use the following approach:
Choose a subset C of S. There are 2^n ways to do this.
Choose a subset B of C. There are 2^|C| ways to do this, where |C| is the size of C.
Choose a subset A of B. There are 2^|B| ways to do this, where |B| is the size of B.
Therefore, the number of ways that A is a subset of B, and B is a subset of C is:
∑_{C⊆S} 2^|C| * 2^|B|
where the sum is over all subsets C of S.
We can simplify this expression by noting that for any fixed subset C of S, the number of ways to choose subsets B and A of C such that A is a subset of B is 2^(|C|+1). This is because we can choose any subset B of C, and then choose any subset A of B. There are 2^(|C|+1) ways to do this.
Therefore, the number of ways that A is a subset of B, and B is a subset of C is:
∑_{C⊆S} 2^(|C|+1)
We can evaluate this sum using the following trick:
∑_{C⊆S} 2^(|C|+1) = 2 * ∑_{C⊆S} 2^|C|
The sum on the right-hand side is just the number of subsets of S, which is 2^n. Therefore, we have:
∑_{C⊆S} 2^(|C|+1) = 2 * 2^n = 2^(n+1)
Finally, the probability that A is a subset of B, and B is a subset of C is:
2^(n+1) / 2^(3n) = 1 / 2^(2n-1)
Therefore, the probability that A is a subset of B, and B is a subset of C is 1 / 2^(2n-1).
