Solution
First Method:
\[
\mathrm{S}=\{1,2,3,4, \ldots, 10,11\}
\]
even no →2,4,6,8,10
odd no →1,3,5,7,9,11
Total subsets $=2^{11}$
No of subsets such that product of numbers is $\mathrm{odd}=2^{6}$
No of subsets such that product of number is even $=2^{11}-2^{6}=2048-64=1984$ required no of subsets $=1984-(\mathrm{n}(\mathrm{B})=1$ \& Product is even $)$
\[
=1984-5=1979 \text { Ans. }
\]
Second Method:
\[
\begin{array}{ll}
\mathrm{n}(\mathrm{~B})=2, & \quad{ }^{11} \mathrm{C}_{2}-{ }^{6} \mathrm{C}_{2} \text { (all are odds) } \\
\mathrm{n}(\mathrm{~B})=3, & { }^{11} \mathrm{C}_{3}-{ }^{6} \mathrm{C}_{3} \text { (all are odds) } \\
\mathrm{n}(\mathrm{~B})=4, & { }^{11} \mathrm{C}_{4}-{ }^{6} \mathrm{C}_{4} \text { (all are odds) } \\
\mathrm{n}(\mathrm{~B})=5, & { }^{11} \mathrm{C}_{5}-{ }^{6} \mathrm{C}_{5} \text { (all are odds) } \\
\mathrm{n}(\mathrm{~B})=6, & { }^{11} \mathrm{C}_{6}-{ }^{6} \mathrm{C}_{6} \text { (all are odds) } \\
\mathrm{n}(\mathrm{~B})=7, & { }^{11} \mathrm{C}_{7}-0 \\
\mathrm{n}(\mathrm{~B})=8 & { }^{11} \mathrm{C}_{8} \\
\mathrm{n}(\mathrm{~B})=9, & { }^{11} \mathrm{C}_{9} \\
\mathrm{n}(\mathrm{~B})=10, & { }^{11} \mathrm{C}_{10} \\
\mathrm{n}(\mathrm{~B})=11, & { }^{11} \mathrm{C}_{11}
\end{array}
\]
required no. of subsets $=\sum_{\mathrm{r}=2}^{11}{ }^{11} \mathrm{C}_{\mathrm{r}}-\sum_{\mathrm{r}=2}^{6}{ }^{6} \mathrm{C}_{\mathrm{r}}$
\[
\begin{aligned}
& =\left(2^{11}-{ }^{11} \mathrm{C}_{0}-{ }^{11} \mathrm{C}_{1}\right)-\left(2^{6}-{ }^{6} \mathrm{C}_{0}-{ }^{6} \mathrm{C}_{1}\right) \\
& =2048-1-11-(64-1-6) \\
& =2048-64-5=1979 \text { Answer }(1979)
\end{aligned}
\]
.
