AMC 10 B

The 16 people can be partitioned into the 4 committees, each of size 4, in
$$\frac{16!}{(4!)^5}$$
ways; four of the $4!$ factors come from permuting the members of the committees and one $4!$ factor comes from permuting the committees. Then there are $4^4$ ways to choose the four chairpersons and $3^4$ ways to choose the four secretaries. This gives a total of
$$\frac{16! \cdot 4^4 \cdot 3^4}{(4!)^5}$$
assignments. The numerator has 1 factor of 3 in each of 15, 12, 6, and 3; it has 2 factors of 3 in 9 and 4 factors in $3^4$, a total of 10. The denominator has 5 factors of 3. Thus $r = 10 - 5 = 5$.

OR

There are $16 \cdot 15$ ways to choose the chairperson and secretary of the first committee, $14 \cdot 13$ ways to choose the chairperson and secretary of the second committee, $12 \cdot 11$ ways to choose the chairperson and secretary of the third committee, and $10 \cdot 9$ ways to choose the chairperson and secretary of the fourth committee. There are then $\frac{8 \cdot 7}{2} = 28$ ways to choose the remaining members of the first committee, $\frac{6 \cdot 5}{2} = 15$ ways to choose the remaining members of the second committee, $\frac{4 \cdot 3}{2} = 6$ ways to choose the remaining members of the third committee, and $\frac{2 \cdot 1}{2} = 1$ way to choose the remaining members of the fourth committee. There are $4!$ ways to account for the fact that the committees are indistinguishable. This gives a total of
$$\frac{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 28 \cdot 15 \cdot 6 \cdot 1}{4!}$$
assignments. There are $1 + 1 + 2 + 1 + 1 = 6$ factors of 3 in the numerator and 1 factor of 3 in the denominator, so $r = 6 - 1 = 5$.