Let S = {1, 2, 3, ...., 100}. The number of non-empty subsets A of S such that the product of elemen

Q: Why do we subtract the empty set? Does the empty set have an even product?

The problem explicitly asks for 'non-empty' subsets, so we must remove the empty set from our count. Mathematically, the product of an empty set is defined as the multiplicative identity, which is $1$. Since $1$ is odd, the empty set would fall into the 'odd product' category, but since we are excluding it from the total count of subsets, we must be careful to subtract it from both the total and the odd-product count to keep our logic consistent.

Q: Why can't we just pick one even number and then any other numbers? That seems more direct.

That approach is a classic trap! If you try to pick one even number and then any combination of the remaining $99$ numbers, you will end up overcounting. For example, if you pick the number $2$ and then the number $4$, you would count that set twice—once when you picked $2$ as your 'first' even number, and again when you picked $4$. Complementary counting is the elegant, foolproof way to avoid this overlap.

Q: Is the product of odd numbers always odd?

Yes, absolutely! The product of any number of odd integers is always odd because there are no factors of $2$ present to make the product even. This fundamental property of parity is the key that unlocks this entire problem. If you have a subset containing only odd numbers, the product is guaranteed to be odd. If you have even one even number, the product is guaranteed to be even. It is a binary, beautiful distinction.

Question

Let $S = \{1, 2, 3, ...., 100\}$. The number of non-empty subsets A of S such that the product of elements in A is even is :

Options:

(A) $2^{50}(2^{50}-1)$

(B) $2^{100}-1$

(C) $2^{50}-1$

(D) $2^{50}+1$

Sigma Percentile · Accepted Answer

### The Anatomy of the Set

We are working with the set $S = \{1, 2, 3, ..., 100\}$. This set contains exactly $50$ odd numbers and $50$ even numbers.

When a problem asks for the existence of "at least one" element satisfying a condition—in this case, at least one even number to ensure an even product—your intuition should immediately turn to **Complementary Counting**.

### The Parity Logic

Counting the "even product" subsets directly is inefficient, as it would require summing combinations for subsets containing one, two, three, up to fifty even numbers. This approach is computationally expensive and prone to error.

Instead, we consider the complement: when is a product **odd**? A product of integers is odd if and only if **every single element** in the subset is odd.

If a subset contains even a single even number, the product is guaranteed to be even. Therefore, the strategy is:
$$	ext{Even-Product Subsets} = 	ext{Total Non-Empty Subsets} - 	ext{Odd-Product Subsets}$$

### The Calculation

First, we calculate the total number of non-empty subsets for a set of $100$ elements. The total number of subsets is $2^{100}$, and excluding the empty set, we have:
$$	ext{Total Non-Empty} = 2^{100} - 1$$

Next, we calculate the number of non-empty subsets that result in an odd product. These subsets must be formed exclusively from the $50$ odd numbers available in $S$. The number of such subsets is $2^{50}$, and excluding the empty set, we have:
$$	ext{Non-Empty Odd-Product} = 2^{50} - 1$$

### The Final Synthesis

Now, we perform the subtraction to find the number of subsets with an even product:
$$	ext{Result} = (2^{100} - 1) - (2^{50} - 1)$$

The constants cancel out perfectly:
$$	ext{Result} = 2^{100} - 2^{50}$$

Factoring out $2^{50}$, we arrive at the final elegant form:
$$\mathbf{2^{50}(2^{50} - 1)}$$

This result demonstrates the power of logical exclusion. By understanding the parity structure of the set, we bypassed complex summation and arrived at the solution through the efficiency of the complement.

Animated Solution for Mathematics - Permutations and Combinations: Let $S = {1, 2, 3, ...., 100}$ . The number of non-empty subsets A of S such that the product of elements in A is even is :

Select Answer:

Visualized Solution

Analyze the Set $S$

Categorize Elements

Condition for Even Product

Condition for Odd Product

Complementary Counting

Total Non-empty Subsets

Odd Product Subsets

Calculate Even Product Subsets

Expand the Expression

Simplify

Final Factorization

The Sigma Insight: Combinations and Selection

The Anatomy of the Set

The Parity Logic

The Calculation

The Final Synthesis

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