JEE Advanced 2025 — Paper 1 — Question 24

Let $\mathrm{A} \rightarrow$ " 0 " appear exactly twice. and $\mathrm{B} \rightarrow " 1 "$ appear exactly twice.

$\therefore \mathrm{A} \cap \mathrm{B} \rightarrow$ " 0 " and " 1 " both appears exactly twice.

$\mathrm{n}(\mathrm{A}) =$ $\underset{\text { placing zero }}{{ }^{6} \mathrm{C}_{2}(1)(2)^{5}}=\frac{6 \times 5}{2} \times 2^{5}=480$

for n (B)

C-I: 1 at first place; Number of ways $={ }^{6} \mathrm{C}_{1}(1)(2)^{5}=192$

C-II : 2 at first place; Number of ways $={ }^{6} \mathrm{C}_{2}(1)(2)^{4}=\frac{6 \times 5}{2} \times 2^{4}=240$

$\mathrm{n}(\mathrm{B})=240+192$

$$\mathrm{n}(\mathrm{~A} \cap \mathrm{~B})={ }_{\text {placing zero }}^{6} \mathrm{C}_{2}(1) \times{ }_{\text {placing 1 }}^{5} \mathrm{C}_{2}(1) \times{ }_{\text {2atrest places 1 }}^{(1)}=\frac{6 \times 5}{2} \times \frac{5 \times 4}{2}=150$$ $$\therefore \mathrm{n}(\mathrm{~A} \cup \mathrm{~B})=\mathrm{n}(\mathrm{~A})+\mathrm{n}(\mathrm{~B})-\mathrm{n}(\mathrm{~A} \cap \mathrm{~B}) =480+(192+240)-150 =762$$