Let S be the set of the first 11 natural numb | vedclass

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(A) The set $S = \{1, 2, 3, . . . , 11\}$ contains $6$ odd numbers $\{1, 3, 5, 7, 9, 11\}$ and $5$ even numbers $\{2, 4, 6, 8, 10\}$.The total number

Vedclass · Accepted Answer

$1979$

Vedclass · Answer

(A) The set $S = \{1, 2, 3, . . . , 11\}$ contains $6$ odd numbers $\{1, 3, 5, 7, 9, 11\}$ and $5$ even numbers $\{2, 4, 6, 8, 10\}$.The total number of subsets of $S$ is $2^{11} = 2048$.$A$ subset $B$ has an odd product if and only if all elements of $B$ are odd. The number of such subsets is $2^6 = 64$.$A$ subset $B$ has an even product if it contains at least one even number. The number of such subsets is $2^{11} - 2^6 = 2048 - 64 = 1984$.The condition $n(B) \ge 2$ excludes subsets with $0$ or $1$ element.Subsets with $0$ elements: $\emptyset$ (product is not defined or considered odd,so it is excluded).Subsets with $1$ element: $\{1\}, \{3\}, \{5\}, \{7\}, \{9\}, \{11\}$ (all odd products) and $\{2\}, \{4\}, \{6\}, \{8\}, \{10\}$ (all even products).We must exclude the $5$ subsets that contain exactly one even number (since their product is even but $n(B) Therefore,the required number of subsets is $1984 - 5 = 1979$.

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

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