2024 AMC 10B Problem 22

Line the $16$ people up in a row ($16!$ ways) and cut the row into four consecutive blocks of $4$. In each block, call the first person the chairperson and the second the secretary; the last two are ordinary members.

This overcounts. Swapping the two ordinary members within a block changes nothing, a factor of $2$ per block, so $2^4$ overall. Reordering the four blocks changes nothing since committees are indistinguishable, a factor of $4!$. Every assignment arises from exactly $2^4 \cdot 4!$ lineups, so the number of assignments is
$$\frac{16!}{2^4 \cdot 4!}.$$

Now count factors of $3$. In $16!$ the multiples of $3$ contribute $\lfloor 16/3 \rfloor = 5$ and the multiple of $9$ contributes one more, for $3^6$. The denominator has $4! = 24 = 3 \cdot 8$, one factor of $3$, and $2^4$ has none. So
$$r = 6 - 1 = 5.$$

The answer is $\boxed{\textbf{(A)}\ 5}$.

"Arrange everyone in a line, then divide out the symmetries you don't care about" produces the count in a form that is already factored, which is exactly what a "power of $3$" question wants. Legendre's formula ($\lfloor n/p \rfloor + \lfloor n/p^2 \rfloor + \cdots$) then reads off the exponent without evaluating anything. Whenever an answer asks for a prime's exponent, keep the count as a product of factorials and small numbers.

Choose committees in order: $\binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4} / 4! = \frac{16!}{(4!)^5}$, then pick chair and secretary in each: $12^4$. Exponent of $3$: $6 - 5 + 4 = 5$.

• Forgetting to divide by 4! for indistinguishable committees, which leaves the exponent at 6 (choice (B)), or counting the chair and secretary as an unordered pair.
• Counting the factors of $3$ in $16!$ as $5$ (missing the extra one from $9$), which gives $r = 4$ and is not even a choice.