The number of 4 -letter words, with or without meaning, which can...

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