2024 AMC 10A Problem 6

Start with the example in the problem. Going from $ABC$ to $CBA$, the pairs $A,B$ and $A,C$ and $B,C$ all have to switch order: $A$ starts before $B$ and ends after it, and so on. That is $3$ pairs, and the example used $3$ swaps. This suggests counting pairs.

1. Count the pairs. In $ABCDEF$, every pair of letters is in alphabetical order. In $FEDCBA$, every pair is in the opposite order. So every pair must switch. With $6$ letters, the number of pairs is
$$5 + 4 + 3 + 2 + 1 = 15$$
($A$ pairs with $5$ letters after it, $B$ with $4$, and so on). This is also $\binom{6}{2} = 15$.
2. One swap flips only one pair. When you swap two letters that are next to each other, only those two letters change order. Every other letter stays on the same side of each of them. So each swap switches exactly one pair.
3. So at least 15 swaps are needed, because $15$ pairs have to switch and each swap handles only one.
4. 15 swaps are enough. Move $A$ to the end one step at a time ($5$ swaps): $BCDEFA$. Then move $B$ right until it sits just before $A$ ($4$ swaps): $CDEFBA$. Then $C$ ($3$ swaps): $DEFCBA$. Then $D$ ($2$ swaps): $EFDCBA$. Then $E$ ($1$ swap): $FEDCBA$. Total: $5 + 4 + 3 + 2 + 1 = 15$.

At least $15$ are needed and $15$ are enough, so the minimum is $15$.

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

When a move can only change a count by $1$, that count tells you the fewest moves you need. Here the count is "how many pairs are still in the wrong order", and each swap of neighbours changes it by exactly $1$. Reversing $n$ letters always takes $1 + 2 + \cdots + (n-1)$ swaps; the example with $3$ letters and $3$ swaps is the small case.

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 $\boxed{\textbf{(D)}\ 15}$.