JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Permutations and Combinations

Let $A$ = event that 0 appears exactly twice, and $B$ = event that 1 appears exactly twice.

For $n(A)$: Choose 2 positions from the last 6 for the two 0's (first position cannot be 0), and fill remaining 4 positions with 1's and 2's:


For $n(B)$: Two cases:

• Case I: First position is 1, choose 1 more position from remaining 6 for the second 1, fill remaining 5 with 0's and 2's: $\left(\genfrac{}{}{0ex}{}{6}{1}\right)\cdot 2^5=6\cdot 32=192$
• Case II: First position is 2, choose 2 positions from remaining 6 for two 1's, fill remaining 4 with 0's and 2's: $\left(\genfrac{}{}{0ex}{}{6}{2}\right)\cdot 2^4=15\cdot 32=480$

Wait, let me recalculate Case II: $\left(\genfrac{}{}{0ex}{}{6}{2}\right)\cdot 2^4=15\cdot 16=240$

So $n(B)=192+240=432$

For $n(A\cap B)$: Both 0 and 1 appear exactly twice. Choose 2 positions from last 6 for 0's, then 2 from remaining 4 for 1's, fill remaining 2 with 2's:


Wait, the source says 150. Let me reconsider: if first digit can be 1 or 2, we need to be more careful.

Actually, following the source solution more carefully: $n(A\cap B)=\left(\genfrac{}{}{0ex}{}{6}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{5}{2}\right)=15\cdot 10=150$

By inclusion-exclusion: