2024 AMC 10B Problem 22

We only need to know how many factors of $3$ the count has, so write the count as a product of simple pieces instead of working out the number.

Step 1: a formula for the count.

1. Line up the $16$ people in a row. There are $16!$ ways.
2. Cut the row into four blocks of $4$: people $1$–$4$, $5$–$8$, $9$–$12$, $13$–$16$. In each block, the first person is chairperson, the second is secretary, and the last two are ordinary members.
3. Every lineup gives a valid set of committees, but many lineups give the same one:
- Swapping the two ordinary members in a block changes nothing. That is $2$ lineups per block, so $2 \cdot 2 \cdot 2 \cdot 2 = 2^4$ in all.
- The committees are indistinguishable, so reordering the four blocks changes nothing. That is $4! = 24$ lineups.
4. So each set of committees is counted $2^4 \cdot 4!$ times, and the number of ways is
$$\frac{16!}{2^4 \cdot 4!}.$$

Step 2: count the factors of $3$.

1. On top, $16! = 1 \cdot 2 \cdot 3 \cdots 16$. The multiples of $3$ are $3, 6, 9, 12, 15$: five of them, each giving one $3$. The number $9 = 3 \cdot 3$ gives one extra. So $16!$ has $6$ factors of $3$.
2. On the bottom, $2^4$ has no $3$s, and $4! = 24 = 3 \cdot 8$ has one.
3. So the count has $6 - 1 = 5$ factors of $3$, and $r = 5$.

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

"Line everyone up, then divide out the orders you do not care about" gives the count as a fraction of factorials and small numbers. That form is exactly what you want when the question asks how many times a prime divides the answer. To count the $3$s in $16!$, count the multiples of $3$, then add one more for each multiple of $9$.

Build the committees directly. Choose $4$ people for a first committee, then $4$ of the remaining $12$, then $4$ of the remaining $8$, and the last $4$ are forced. Divide by $4!$ because the committees have no order. This comes to $\frac{16!}{(4!)^5}$. Then choose a chair ($4$ ways) and a secretary ($3$ ways) in each committee: $12^4$ more. Count the $3$s: $16!$ has $6$, $(4!)^5$ has $5$, and $12^4$ has $4$. So $r = 6 - 5 + 4 = 5$.

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 $\boxed{\textbf{(A)}\ 5}$.