JEE Main 2026 — Permutation Combination Question with Solution

Letters: P(3), Q(2), R(2), S(1), T(1), U(1), V(1) — 7 distinct letters.
All distinct: $\left(\genfrac{}{}{0ex}{}{7}{4}\right)\times 4!=840$.
One pair + 2 distinct: choose repeating letter from $\{P,Q,R\}$ (3 ways), choose 2 from remaining 6 ($\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$), arrange $=4!/2!=12$. Total $=3\times 15\times 12=540$.
Two pairs: choose 2 from $\{P,Q,R\}$ ($\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$), arrange $=4!/(2!2!)=6$. Total $=18$.
One triple + 1 distinct: only P can repeat 3 times, choose 1 from 6, arrange $=4!/3!=4$. Total $=24$.
Answer $=840+540+18+24=1422$.