If all the words with or without meaning made using all the letters of the word ``KANPUR'' are arranged as in a dictionary, then the word at $440^{\text{th}}$ position in this arrangement is:
Step-by-Step Solution
Key Concept: Sort the letters: A, K, N, P, R, U. At each position, count how many words start with each smaller letter ($=5!,4!,3!,\dots$) and chip away at the target rank.
Sorted letters: A, K, N, P, R, U. Total $6!=720$ words.
Block sizes of $5!=120$:
A$\cdots$: positions $1$–$120$;\quad K$\cdots$: $121$–$240$;\quad N$\cdots$: $241$–$360$;\quad P$\cdots$: $361$–$480$.
$440$ is in P$\cdots$ block, offset $440-360=80$.
Within P, sorted remaining = A,K,N,R,U. Block sizes of $4!=24$:
PA: $1$–$24$;\quad PK: $25$–$48$;\quad PN: $49$–$72$;\quad PR: $73$–$96$.
$80$ is in PR, offset $80-72=8$.
Within PR, sorted remaining = A,K,N,U. Block sizes of $3!=6$:
PRA: $1$–$6$;\quad PRK: $7$–$12$.
$8$ is in PRK, offset $8-6=2$.
Within PRK, sorted remaining = A,N,U. Block sizes of $2!=2$:
PRKA: $1$–$2$;\quad PRKN: $3$–$4$;\quad PRKU: $5$–$6$.
$2$ is in PRKA. Within PRKA, sorted remaining N,U gives PRKANU ($1$) and PRKAUN ($2$).
Hence the $440^{\text{th}}$ word is $\boxed{\text{PRKAUN}}.$
Correct Answer: 3