JEE Advanced 2025 Paper 1 · Q10 · Permutations & Combinations

Question

Let $S$ be the set of all seven-digit numbers formed using the digits $0, 1$, and $2$ (so the leading digit must be $1$ or $2$). Find the number of elements $x$ in $S$ such that AT LEAST ONE of the digits $0$ and $1$ appears exactly twice in $x$.

SolveKar AI · Accepted Answer

Use inclusion–exclusion. Let $A = \{x: 	ext{digit } 0 	ext{ appears exactly twice}\}$, $B = \{x: 	ext{digit } 1 	ext{ appears exactly twice}\}$. Answer $= |A| + |B| - |A \cap B|$. $|A|$: choose $2$ positions for the two $0$s among $7$, but the leading position cannot be $0$. So choose $2$ of the last $6$ positions: $\binom{6}{2}=15$. The remaining $5$ positions are filled with $1$s and $2$s freely (with no constraint that the leading digit is non-zero, since it is among these $5$): $2^5 = 32$ ways. $|A| = 15 \cdot 32 = 480$. $|B|$: choose $2$ of $7$ positions for the two $1$s: $\binom{7}{2} = 21$. Remaining $5$ positions are filled with $0$s and $2$s, but leading digit cannot be $0$. If leading digit is among the chosen-$1$ positions (cases $\binom{6}{1}=6$ with leading $=1$): then $5$ remaining slots all free: $2^5=32$. If leading digit not among the $1$-positions (cases $\binom{6}{2}=15$): then leading must be $2$, the other $4$ free: $2^4 = 16$. $|B| = 6 \cdot 32 + 15 \cdot 16 = 192 + 240 = 432$. $|A \cap B|$: two $0$s and two $1$s among $7$ digits, remaining three are $2$s. $0$s must avoid leading position. Choose $2$ of last $6$ for $0$s: $15$. Choose $2$ of re…[truncated]

JEE Advanced 2025 Paper 1 · Q10 · Permutations & Combinations

Solution