2024 AMC 10A Problem 6

Look at pairs of letters. In $ABCDEF$ every pair is in alphabetical order; in $FEDCBA$ every pair is reversed. There are $\binom{6}{2} = 15$ pairs.

A swap of two adjacent letters changes the relative order of exactly one pair (the two letters swapped) and leaves every other pair alone. So each swap fixes at most one of the $15$ pairs, and at least $15$ swaps are needed.

Fifteen swaps suffice: move $A$ to the right end with $5$ swaps ($BCDEFA$), then $B$ to the second-to-last spot with $4$ swaps ($CDEFBA$), then $C$ with $3$, $D$ with $2$, $E$ with $1$: total $5+4+3+2+1 = 15$, ending at $FEDCBA$.

The answer is $\boxed{\textbf{(D)}\ 15}$.

The number of out-of-order pairs (inversions) changes by exactly $1$ per adjacent swap, so it is a monovariant that gives a lower bound, and the greedy "bubble" construction shows the bound is attained. Reversing $n$ letters always takes $\binom{n}{2}$ swaps; the given example $ABC \to CBA$ with $3 = \binom{3}{2}$ swaps is the small case of this pattern.

• Guessing 6 (one swap per letter) or 24 from a factorial instead of counting the pairs whose order must change.
• Finding a $15$-swap sequence but not arguing why fewer swaps cannot work.