JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Permutations and Combinations

The word EXAMINATION has $11$ letters, with A, I and N each appearing twice and E, X, M, T, O once, so there are $8$ distinct letters. For four all-different letters, choose and arrange $4$ of the $8$: ${}^8{P}_4=1680$. For two alike and two different, pick the repeated letter from the three available, two more from the remaining $7$ distinct letters, and arrange: ${}^3{C}_1\cdot {}^7{C}_2\cdot \frac{4!}{2!}=3\cdot 21\cdot 12=756$. For two pairs, pick two repeated letters from the three and arrange: ${}^3{C}_2\cdot \frac{4!}{2!2!}=3\cdot 6=18$. Altogether $1680+756+18=2454$.