Let S be the set of the first 11 natural numbers.

Question

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

Answer: 1979

Accepted Answer

Let $S = \{1, 2, \ldots, 11\}$. The product of all elements of $B$ is even iff $B$ contains at least one even number.

- Evens in $S$: 5 elements, namely $2, 4, 6, 8, 10$.
- Odds in $S$: 6 elements.

Number of subsets containing at least one even element:
$2^{11} - 2^{6} = 2048 - 64 = 1984$
(subtracting the subsets made entirely of odd numbers).

Now enforce $n(B) \geq 2$: subtract the 5 singleton subsets $\{2\}, \{4\}, \{6\}, \{8\}, \{10\}$ (each has exactly one, even, element and size 1).
$1984 - 5 = 1979$

Solutions · 1