The number of $6$-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: Split by the number of distinct letters used ($1,2,$ or $3$ — never $4+$, since $4\cdot 2>6$). For each, choose the letters and distribute $6$ slots into multiplicity vectors with each part $\ge 2.$
\textbf{$1$ distinct letter:} pick letter from $5$, $\binom{5}{1}=5$ words.
\textbf{$2$ distinct letters:} pick pair $\binom{5}{2}=10$. Multiplicity vectors with each $\ge 2$ and sum $6$: $(2,4),(3,3),(4,2)$, contributing $\dfrac{6!}{2!4!}+\dfrac{6!}{3!3!}+\dfrac{6!}{4!2!}=15+20+15=50.$ Total $=10\cdot 50=500.$
\textbf{$3$ distinct letters:} each must appear exactly $2$ times. Pick $\binom{5}{3}=10$. Arrangements $=\dfrac{6!}{2!2!2!}=90.$ Total $=10\cdot 90=900.$
\textbf{$4+$ distinct letters:} impossible ($\ge 4\cdot 2=8>6$).
Grand total: $5+500+900=1405.$
Correct Answer: 1405