JEE Main 2026 — 22 January, Evening Shift — Question 25

A $\{1,2,3 \ldots \ldots 11\}$

$\therefore \mathrm{n}(\mathrm{B}) \geq 2$ & product of all elements in B is even

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

$\mathrm{n}(\mathrm{B})=3 \Rightarrow{ }^{11} \mathrm{C}_{3}-{ }^{6} \mathrm{C}_{3}$

$\mathrm{n}(\mathrm{B})=4 \Rightarrow{ }^{11} \mathrm{C}_{4}-{ }^{6} \mathrm{C}_{4}$

$\mathrm{n}(\mathrm{B})=5 \Rightarrow{ }^{11} \mathrm{C}_{5}-{ }^{6} \mathrm{C}_{5}$

$\mathrm{n}(\mathrm{B})=6 \Rightarrow{ }^{11} \mathrm{C}_{6}-{ }^{6} \mathrm{C}_{6}$

$\mathrm{n}(\mathrm{B})=7 \Rightarrow{ }^{11} \mathrm{C}_{7}$ : : $n(B)=11 \Rightarrow{ }^{11} C_{11}$

∴ number of set $\mathrm{B} \Rightarrow \sum_{\mathrm{r}=2}^{11}{ }^{11} \mathrm{C}_{\mathrm{r}}-\sum_{\mathrm{r}=2}^{6}{ }^{6} \mathrm{C}_{\mathrm{r}}$

$=2^{11}-(12)-\left(2^{6}-7\right)$

$=2048-64-5$ = 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 $=5$

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