Permutations and Combinations Part 2

Concept:Count 4-letter words by cases based on letter repetition patterns using permutations with identical objects.Explanation:Letters: $P (3)$, $Q,R (2)$ each, $S,T,U,V (1)$ each. Total distinct letters: $7$.Consider four patterns:1. All four distinct: Choose $4$ distinct letters from $7$: $\binom{7}{4}$ ways. Arrange them: $4! = 24$. So $35 \times 24 = 840$ words.2. One letter repeated twice, two singles (AABC): Choose repeated letter from $\{P,Q,R\}$: $\binom{3}{1}$ ways. Choose two singles from remaining $6$ distinct letters: $\binom{6}{2}$ ways. Arrange: $\frac{4!}{2!} = 12$. So $3 \times 15 \times 12 = 540$ words.3. Two letters each repeated twice (AABB): Choose two letters from $\{P,Q,R\}$: $\binom{3}{2}$ ways. Arrange: $\frac{4!}{2!\,2!} = 6$. So $3 \times 6 = 18$ words.4. One letter repeated thrice, one single (AAAB): Only $P$ can repeat thrice: $1$ way. Choose single from remaining $6$ letters: $\binom{6}{1}$ ways. Arrange: $\frac{4!}{3!} = 4$. So $1 \times 6 \times 4 = 24$ words.Total: $840 + 540 + 18 + 24 = 1422$ words.Answer:$\boxed{1422}$ (Option A)