JEE Advanced 2025 · Paper 1, Mathematics question 10

Let $A$ = "0 appears exactly twice" and $B$ = "1 appears exactly twice". The first digit is 1 or 2.

$|A|$. Put the two 0s in the last 6 places (${}^6C_2=15$ ways). The other 5 digits, including the first, are 1 or 2 ($2^5$ ways):
$$|A|=15\cdot32=480.$$

$|B|$.

• First digit 1: one more 1 in the last 6 (6 ways), the rest from $\{0,2\}$ ($2^5$): $192$.
• First digit 2: two 1s in the last 6 (15 ways), the rest from $\{0,2\}$ ($2^4$): $240$.
• Total: $432$.

$|A\cap B|$.

• First digit 1: the last 6 hold one 1, two 0s and three 2s: $\frac{6!}{1!\,2!\,3!}=60$.
• First digit 2: the last 6 hold two 1s, two 0s and two 2s: $\frac{6!}{2!\,2!\,2!}=90$.
• Total: $150$.

$$|A\cup B|=480+432-150=762.$$