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Question 33 — JEE Main 22 Jan 2026 S Ii

Question

Let S be the set of the first 11 natural numbers. Then the number of elements in $\mathrm{A}=\{\mathrm{B} \subseteq \mathrm{S}: \mathrm{n}(\mathrm{B}) \geq 2$ and the product of all elements of B is even $\}$ is $\_\_\_\_$
[JEE MAIN 22 Jan 2026 S II]

Solution

First Method:

$\mathrm{S}=\{1,2,3,4, \ldots, 10,11\}$
even no $\rightarrow 2,4,6,8,10$
odd no $\rightarrow 1,3,5,7,9,11$
Total subsets $=2^{11}$
No of subsets such that product of numbers is 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, & { }^{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 $=\sum_{\mathrm{r}=2}^{11}{ }^{11} \mathrm{C}_{\mathrm{r}}-\sum_{\mathrm{r}=2}^{6}{ }^{6} \mathrm{C}_{\mathrm{r}}$
$=\left(2^{11}-{ }^{11} \mathrm{C}_{0}-{ }^{11} \mathrm{C}_{1}\right)-\left(2^{6}-{ }^{6} \mathrm{C}_{0}-{ }^{6} \mathrm{C}_{1}\right)$
$=2048-1-11-(64-1-6)$
$=2048-64-5=1979$ Answer(1979)