JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Permutations and Combinations

The nine-digit numbers use the digits $a,a,a,b,b,b,c,c,c$, giving $\frac{9!}{3!3!3!}=1680$ distinct numbers. Three consecutive digits are in arithmetic progression when they read $abc$ or $cba$. Counting the distinct numbers that contain at least one such run, and subtracting through inclusion-exclusion those counted more than once because they contain two or three runs, leaves $944$ numbers. The placement count $7\times 2\times 90=1260$ instead counts runs with multiplicity and so overstates the total. Hence $944$ such numbers can be formed.