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 _______. [2026]

Question

Let S be the set of the first 11 natural numbers. Then the number of elements in

$A = {B \subseteq S : n (B) \geq 2 and the product of all elements of B is even}$ is _______. [2026]

Smart Achievers · Accepted Answer

(1979)

$A = {1, 2, 3, \dots, 11}$

$∴$ $n (B) \geq 2 and product of all elements in B is even$

$Case (i):$

$n (B) = 2 \Rightarrow {}^{11}{C_{2}} - {}^{6}{C_{2}}$

$n (B) = 3 \Rightarrow {}^{11}{C_{3}} - {}^{6}{C_{3}}$

$n (B) = 4 \Rightarrow {}^{11}{C_{4}} - {}^{6}{C_{4}}$

$n (B) = 5 \Rightarrow {}^{11}{C_{5}} - {}^{6}{C_{5}}$

$n (B) = 6 \Rightarrow {}^{11}{C_{6}} - {}^{6}{C_{6}}$

$n (B) = 7 \Rightarrow {}^{11}{C_{7}}$

$⋮$

$n (B) = 11 \Rightarrow {}^{11}{C_{11}}$

$∴$ $number of set B = \sum_{r = 2}^{11} {}^{11}{C_{r}} - \sum_{r = 2}^{6} {}^{6}{C_{r}}$

$= 2^{11} - (1 + 11) - (2^{6} - 7)$

$= 2048 - 64 - 5$

$= 1979$

$Alternate Solution:$

$Total subsets = 2^{11}$

$No. of subsets having odd terms only = 2^{6}$

$No. of subsets having one term only & also having even terms = 5$

$Required ways = 2^{11} - 2^{6} - 5 = 1979$

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