Key Concepts and Formulas

• Probability Definition: The probability of an event occurring is the ratio of the number of favorable outcomes to the total number of possible outcomes. $P(\text{Event}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$
• Number of Subsets: For a set with $n$ distinct elements, the total number of distinct subsets (including the empty set and the set itself) is $2^n$.
• Sum of First $n$ Natural Numbers: The sum of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
• Complement Principle for Set Sums: If $B$ is a subset of $S$, then the sum of elements in $S$ is equal to the sum of elements in $B$ plus the sum of elements in its complement $S \setminus B$. $\sum_{x \in S} x = \sum_{x \in B} x + \sum_{y \in S \setminus B} y$

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Step-by-Step Solution

Step 1: Determine the Total Number of Possible Subsets

We are given the set $S = \{1, 2, \ldots, 20\}$. This set contains $n=20$ distinct elements. To form any subset of $S$, we consider each element individually. For each of the 20 elements, there are two independent choices: either the element is included in the subset, or it is excluded. Since there are 20 elements, and 2 choices for each, the total number of distinct subsets that can be formed from $S$ is $2 \times 2 \times \ldots \times 2$ (20 times). Therefore, the total number of possible outcomes (total distinct subsets of $S$) is: $\text{Total Number of Subsets} = 2^{20}$

Step 2: Determine the Number of "Nice" Subsets (Favorable Outcomes) using the Complement Principle

A subset $B$ of $S$ is defined as "nice" if the sum of its elements is 203. Directly trying to find all subsets that sum to 203 would be very complex and prone to errors. Instead, we employ a clever strategy using the complement principle.

First, let's calculate the sum of all elements in the set $S$: $\sum_{k=1}^{20} k = \frac{20 \times (20+1)}{2} = \frac{20 \times 21}{2} = 10 \times 21 = 210$ Now, consider a "nice" subset $B$. Its elements sum to 203. Let $S \setminus B$ be the complement of $B$ in $S$. The elements in $S \setminus B$ are those elements from $S$ that are not in $B$. Using the complement principle for set sums: $\sum_{x \in S} x = \sum_{x \in B} x + \sum_{y \in S \setminus B} y$ Substitute the known values: $210 = 203 + \sum_{y \in S \setminus B} y$ Solving for the sum of elements in the complement $S \setminus B$: $\sum_{y \in S \setminus B} y = 210 - 203 = 7$ This is a powerful transformation! Instead of finding subsets that sum to a large number (203), we now need to find subsets of $S$ whose elements sum to a much smaller number (7). Each unique subset $S \setminus B$ that sums to 7 uniquely defines a "nice" subset $B$. Thus, the number of "nice" subsets is equal to the number of subsets of $S$ whose elements sum to 7.

Step 3: Systematically List Subsets of S whose Elements Sum to 7

We need to find all distinct subsets $X \subseteq S$ such that $\sum_{x \in X} x = 7$. Remember that elements within a set must be distinct.

1. Subsets with one element:

   • The only subset containing a single element that sums to 7 is $\{7\}$.
   • (1 subset)

2. Subsets with two elements: Let the elements be $a$ and $b$, with $a, b \in S$ and $a < b$ to ensure distinctness and avoid duplicates. We need $a+b=7$.

   • If $a=1$, then $b=6$. So, $\{1,6\}$.
   • If $a=2$, then $b=5$. So, $\{2,5\}$.
   • If $a=3$, then $b=4$. So, $\{3,4\}$.
   • (3 subsets)

3. Subsets with three elements: Let the elements be $a, b, c$, with $a, b, c \in S$ and $a < b < c$. We need $a+b+c=7$.

   • The smallest possible sum of three distinct elements from $S$ is $1+2+3 = 6$. Since $6 \le 7$, it's possible.
   • If we choose $a=1$ and $b=2$: Then $1+2+c=7 \implies 3+c=7 \implies c=4$. So, $\{1,2,4\}$.
   • If we choose $a=1$ and $b=3$: Then $1+3+c=7 \implies 4+c=7 \implies c=3$. This would mean $c=b=3$, which violates $b<c$ and distinctness. So, $\{1,3,3\}$ is not a valid set.
   • Any other choices for $a,b$ (e.g., $a=1, b=4$) would make $c$ too small or violate distinctness, or the sum would exceed 7.
   • (1 subset)

4. Subsets with four or more elements:

   • The smallest possible sum for four distinct elements from $S$ is $1+2+3+4 = 10$.
   • Since $10 > 7$, it is impossible to form a sum of 7 using four or more distinct elements from $S$.

By systematically listing, we found the following 5 distinct subsets of $S$ whose elements sum to 7:

1. $\{7\}$
2. $\{1,6\}$
3. $\{2,5\}$
4. $\{3,4\}$
5. $\{1,2,4\}$

Each of these 5 subsets corresponds to a unique $S \setminus B$, and thus to a unique "nice" subset $B$. Therefore, the number of favorable outcomes (number of "nice" subsets) is 5.

Step 4: Calculate the Probability

Using the probability formula: $P(\text{Nice Subset}) = \frac{\text{Number of Nice Subsets}}{\text{Total Number of Subsets}}$ $P(\text{Nice Subset}) = \frac{5}{2^{20}}$

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Common Mistakes & Tips

• Ignoring the Complement Principle: A common mistake is to attempt to directly list subsets that sum to 203, which is computationally intensive and error-prone. Always consider the complementary approach when dealing with sums that are large or close to the total sum.
• Violation of Distinctness: When listing subsets, remember that elements within a set must be distinct. Forgetting this can lead to incorrect counts (e.g., counting $\{1,3,3\}$ as a valid subset).
• Incomplete or Unsystematic Enumeration: When finding subsets that sum to a small number, ensure you list them systematically (e.g., by number of elements, then by smallest element) to avoid missing any combinations or double-counting.

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Summary

We first determined that the total number of possible subsets for a set with 20 elements is $2^{20}$. To find the number of "nice" subsets (those whose elements sum to 203), we used the complement principle. The total sum of elements in $S$ is 210. If a subset $B$ sums to 203, its complement $S \setminus B$ must sum to $210 - 203 = 7$. By systematically listing all subsets of $S$ that sum to 7, we found there are 5 such subsets: $\{7\}$, $\{1,6\}$, $\{2,5\}$, $\{3,4\}$, and $\{1,2,4\}$. Therefore, there are 5 "nice" subsets. The probability is then the ratio of favorable outcomes to total outcomes, which is $\frac{5}{2^{20}}$.

The final answer is $\boxed{\frac{5}{2^{20}}}$, which corresponds to option (A).