1989 AHSME Problems/Problem 24

Problem

Five people are sitting at a round table. Let $f\geq 0$ be the number of people sitting next to at least 1 female and $m\geq0$ be the number of people sitting next to at least one male. The number of possible ordered pairs $(f,m)$ is

$\mathrm{(A) \ 7 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 9 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 11 }$

Solution

Suppose there are more men than women; then there are between zero and two women.

If there are no women, the pair is $(0,5)$ . If there is one woman, the pair is $(2,5)$ .

If there are two women, there are two arrangements: one in which they are together, and one in which they are apart, giving the pairs $(4,5)$ and $(3,4)$ .

All four pairs are asymmetrical; therefore by symmetry there are eight pairs altogether, so $\rm{(B)}$ .

Solution 2

Denote $T_n$ as the number of such pairs for $n$ people. Then for $T_{n-1}$ , when we add an extra spot, we can either have a male or female giving two options. Note that these two options however double the value of $T_{n-1}$ . Now if we note that $T_2=1$ , we have that $T_5=8$ , so that the answer is $\rm{(B)8}$ .

1989 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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1989 AHSME Problems/Problem 24

Contents

Problem

Solution

Solution 2

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