JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Permutations and Combinations

Use the inclusion-exclusion principle. Total permutations of 7 digits = $7!$.

Let $A$ = permutations containing the string 153 (treat 153 as a single block): $n(A)=5!$.

Let $B$ = permutations containing the string 2467 (treat 2467 as a single block): $n(B)=4!$.

For $A\cap B$, both strings must appear, which gives the arrangement 1532467 or 2467153, so $n(A\cap B)=2!$.

By inclusion-exclusion: $n(A\cup B)=5!+4!-2!=120+24-2=142$.

Permutations with neither string = $7!-142=5040-142=4898$.