Let S be the set of the first 11 natural numbers. Then the number of elements in A = {B ⊆ S : n(B) ≥

Q: Wait, why do we subtract the empty set and singletons? The problem just says $n(B) \ge 2$.

That is a great observation! The condition $n(B) \ge 2$ explicitly excludes any subset with zero elements (the empty set) and any subset with exactly one element (the singletons). Since the total number of subsets $2^{11}$ includes these cases, we must manually remove them to satisfy the constraint.

Q: Why is the product odd only when all elements are odd? Can't it be odd otherwise?

Think about the property of multiplication. If you multiply any number by an even number, the result is always even. Therefore, for a product to be odd, every single factor in that product must be odd. If even one even number is included in the subset, the entire product becomes even. This is the core logic behind our complementary counting strategy!

Q: Why can't we just count the subsets with an even product directly?

You absolutely could, but it would be a combinatorial nightmare! You would have to sum the cases where the subset contains one even number, two even numbers, three even numbers, and so on, while also ensuring the total size is at least two. Complementary counting is the elegant shortcut that saves you from that tedious work.

Question

Let $S$ be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \ge 2 	ext{ and the product of all elements of } B 	ext{ is even }\}$ is ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Universe of Possibilities

Every combinatorial problem begins by defining the total universe. Given the set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$, the total number of possible subsets is determined by the fact that each of the $11$ elements has two choices: it is either in the subset or it is not.

Thus, the total number of subsets is:
$$2^{11} = 2048$$

However, we must satisfy the constraint $n(B) \ge 2$. This requires us to exclude the empty set (which has $0$ elements) and all singleton sets (which have $1$ element).

There is $1$ empty set and $11$ singleton sets. Subtracting these $12$ invalid subsets from the total gives:
$$2048 - 12 = 2036$$
These are the subsets that satisfy our size constraint.

### The Even Product Trap

We now face the second condition: the product of the elements must be even. Counting this directly is inefficient, as it requires considering subsets with varying counts of even numbers.

Instead, we use **complementary counting**. The opposite of an even product is an odd product. A product is odd if and only if every single element in the subset is odd.

### The Odd Complement

Let us isolate the odd numbers in our set $S$. They are $\{1, 3, 5, 7, 9, 11\}$, totaling $6$ odd numbers.

If we form a subset using only these numbers, the product will always be odd. The number of such subsets is:
$$2^6 = 64$$

We must still respect our size constraint of $n(B) \ge 2$. We remove the empty set and the $6$ singleton sets formed exclusively from these odd numbers:
$$64 - (1 + 6) = 64 - 7 = 57$$
There are $57$ subsets that have an odd product and satisfy the size constraint.

### Final Calculation

We now subtract the number of subsets with an odd product from the total number of subsets that satisfy the size constraint.

The calculation is:
$$2036 - 57 = 1979$$

By using the complement, we transformed a complex counting problem into a simple subtraction. The total number of subsets $B$ such that $n(B) \ge 2$ and the product of all elements in $B$ is even is **1979**.

Animated Solution for Mathematics - Permutations and Combinations: Let $S$ be the set of the first 11 natural numbers. Then the number of elements in $A = {B \subseteq S : n (B) \geq 2 and the product of all elements of B is even}$ is ......... .

Enter Numerical Value:

Visualized Solution

Defining the Set $S$

Total Subsets of $S$

Applying the Size Constraint $n (B) \geq 2$

The 'Even Product' Strategy

Identifying Odd Numbers

Calculating Odd Product Subsets

Final Calculation

The Sigma Insight: Combinations and Selection

Analyzing the Universe of Possibilities

The Even Product Trap

The Odd Complement

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Defining the Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Total Subsets of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Applying the Size Constraint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(B)≥2

The 'Even Product' Strategy

Identifying Odd Numbers

Calculating Odd Product Subsets

Final Calculation

The Sigma Insight: Combinations and Selection

Analyzing the Universe of Possibilities

The Even Product Trap

The Odd Complement

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 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 :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={1,2,3,4,5,6,9}. Then the number of elements in the set T={A⊆S:A=ϕ and the sum of all the elements of A is not a multiple of 3} is .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={1,2,3,4}. The total number of unordered pairs of disjoint subsets of S is equal to

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Defining the Set $S$

Total Subsets of $S$

Applying the Size Constraint $n (B) \geq 2$

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