JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Permutations and Combinations

Seven positions are filled from five digits with each used at least once, so the two surplus slots make either one digit occur three times or two digits occur twice. If one digit repeats thrice: choose it in $\left(\genfrac{}{}{0ex}{}{5}{1}\right)=5$ ways and arrange in $\frac{7!}{3!}=840$ ways, giving $5\cdot 840=4200$. If two digits each repeat twice: choose them in $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ ways and arrange in $\frac{7!}{2!2!}=1260$ ways, giving $10\cdot 1260=12600$. The total number of such seven-digit numbers is $4200+12600=16800$.