How many elements are in the set S?

The set S contains the first 11 natural numbers, so n(S) = 11.

How many even and odd numbers are in the first 11 natural numbers?

There are 5 even numbers {2, 4, 6, 8, 10} and 6 odd numbers {1, 3, 5, 7, 9, 11}.

What is the total number of subsets of a set with 11 elements?

The total number of subsets is 2^11 = 2048.

How many subsets of S have only odd elements?

Since there are 6 odd numbers, there are 2^6 = 64 such subsets.

Which sets must be subtracted to satisfy n(B) ≥ 2?

We must subtract the empty set and all singleton sets that have an even element.

What is the final numerical answer?

The total number of such subsets is 1979.

Is the empty set considered to have an even product?

In this context, the product of an empty set is usually considered the identity (1), which is odd, but the condition n(B) ≥ 2 excludes it regardless.

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

Q: What is the condition for the product of elements in a set to be even?

A product of integers is even if and only if at least one element in the set is even.

Q: How do we find subsets with an even product?

Subsets with an even product = (Total subsets) - (Subsets with only odd elements).

Q: What does n(B) ≥ 2 mean?

It means the subset must contain at least two elements, excluding empty sets and singleton sets.

Let S be the set of the first 11 natural numbers. Then the number of elements in A = {B ⊆ S : n(B) ≥ 2 and the product of all elements of B is even} is | JEE Main Mathematics

JEERANKERS

Math Combinatorics

Q Numerical Sets & Counting

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

✅ Correct Answer

1979

✓

Solution Steps

Identify the set elements

The set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$.

Number of elements $n(S) = 11$.

Even numbers in $S = \{2, 4, 6, 8, 10\}$ (Total = 5).

Odd numbers in $S = \{1, 3, 5, 7, 9, 11\}$ (Total = 6).

Condition for even product

A product of integers is even if and only if **at least one** element in the set is even.

Therefore, we need to find subsets $B$ that contain at least one even number from the 5 available even numbers.

Total subsets with even product

Total possible subsets of $S = 2^{11} = 2048$.

Subsets containing only odd numbers = $2^6 = 64$ (since there are 6 odd numbers).

Subsets with an even product = (Total Subsets) – (Subsets with only odd numbers) = $2048 – 64 = 1984$.

Apply the condition $n(B) \geq 2$

The current count (1984) includes sets with $n(B) < 2$, i.e., empty sets and singleton sets.

1. **Empty set ($\phi$):** The empty set has an odd product (by convention, the identity 1) or no product. In either case, it doesn’t contain an even number, so it was already excluded in the $2048 – 64$ calculation.

2. **Singleton sets $\{x\}$:** We must remove sets where $x$ is even but $n(B) = 1$. The even singleton sets are $\{2\}, \{4\}, \{6\}, \{8\}, \{10\}$. There are 5 such sets.

Final Calculation

Number of elements in $A = (\text{Subsets with even product}) – (\text{Singleton even sets})$.

$1984 – 5 = 1979$.

Answer: 1979

📚

Theory

1. Subsets and Power Sets

For any finite set $S$ with $n$ elements, the total number of possible subsets (the power set) is given by $2^n$. This includes the empty set and the set itself. In combinatorics problems, we often use the complement method: finding the total number of outcomes and subtracting those that do not meet our criteria. This is particularly useful when the condition is “at least one,” as calculating the total and subtracting the “none” case is computationally simpler.

2. Parity of Products

The product of a set of integers is even if at least one multiplier is even. Conversely, the product is odd if and only if all multipliers are odd. This fundamental property of parity allows us to partition the problem based on the number of even elements. If a set has $e$ even elements and $o$ odd elements, any subset formed exclusively from the $o$ elements will have an odd product. Any subset containing at least one of the $e$ elements will result in an even product.

3. Constraints on Set Cardinality

The notation $n(B) \geq 2$ imposes a size constraint on the subsets. When counting subsets with specific properties, we must often “filter” our results. For a set with $n$ elements, there is 1 empty set ($n=0$) and $n$ singleton sets ($n=1$). In this problem, we must ensure that the subsets we count not only have an even product but also have a cardinality of at least 2. This requires careful subtraction of the “even” singleton sets.

4. Systematic Counting in Combinatorics

The “First $n$ Natural Numbers” is a common setup in JEE problems. For any $n$, there are $\lfloor n/2 \rfloor$ even numbers and $\lceil n/2 \rceil$ odd numbers. For $n=11$, we have 5 even and 6 odd. Understanding this distribution is the first step. The second step is applying the binomial theorem logic: $\sum_{k=0}^n \binom{n}{k} = 2^n$. This problem essentially asks for a sum of specific combinations, but using $2^n$ logic is the most efficient path to the solution.

❓

FAQs

Why did we subtract 64 from 2048 first?

⌄

64 represents the number of subsets that contain ONLY odd numbers. Since an even product requires at least one even number, we subtract all-odd subsets from the total.

Does the empty set have an even product?

⌄

No, the product of elements in an empty set is typically defined as 1 (the multiplicative identity), which is odd. Thus it’s already excluded from the “even product” count.

What does n(B) ≥ 2 mean?

⌄

It means the subset must have 2 or more elements. We must exclude the empty set (0 elements) and singleton sets (1 element).

How many singleton sets have an even product?

⌄

Only the singleton sets containing an even number: {2}, {4}, {6}, {8}, and {10}. There are 5 such sets.

What if the question asked for an ODD product?

⌄

Then we would look at subsets of only the 6 odd numbers ($2^6 = 64$) and subtract the empty set and 6 odd singleton sets to satisfy $n(B) \geq 2$.

Is zero a natural number in this context?

⌄

In JEE Mathematics, natural numbers ($N$) typically start from 1. So $S = \{1, 2, …, 11\}$.

Why is 2^11 used?

⌄

Every element in a set of 11 has two choices: either it is in the subset or it is not. $2 \times 2 \times … (11 \text{ times}) = 2^{11}$.

Can we solve this using combinations?

⌄

Yes, but it’s much longer: $\sum_{k=2}^{11} (\binom{11}{k} – \binom{6}{k})$. This is equivalent to the power set method used here.

Are there any special cases for product of singletons?

⌄

The product of a singleton set $\{x\}$ is simply $x$ itself. So the product is even if $x$ is even.

How do I verify 2^11 quickly?

⌄

Remember $2^{10} = 1024$. So $2^{11} = 1024 \times 2 = 2048$.

🔗