$26 \cdot 17$ For all $m, n$ satisfying $1 \leqslant n \leqslant m \leqslant 5$, the polar equation
$$
\rho=\frac{1}{1-C_{m}^{n} \cos \theta}
$$

represents the number of different hyperbolas is
(A) 15 .
(B) 10 .
(C) 7 .
(D) 6 .
(China High School Mathematics League, 1984)
[Solution] $\because$ The equation $\rho=\frac{1}{\left(1-C_{m}^{n} \cos \theta\right)}$ represents a hyperbola,
$\therefore \quad C_{m}^{n}>1$. And $1 \leqslant n \leqslant m \leqslant 5$,
$\therefore C_{m}^{n}$ can be expressed as $C_{5}^{1}, C_{5}^{2}, C_{5}^{3}, C_{5}^{4}, C_{5}^{5}, C_{4}^{1}, C_{4}^{2}, C_{4}^{3}, C_{4}^{4}, C_{3}^{1}, C_{3}^{2}, C_{3}^{3}$, $C_{2}^{1}, C_{2}^{2}, C_{1}^{1}$, a total of fifteen expressions.

But $C_{1}^{1}=C_{2}^{2}=C_{3}^{3}=C_{4}^{4}=C_{5}^{5}=1$ do not meet the requirements, so these five expressions should be removed.

Also, $C_{5}^{1}=C_{5}^{4}, C_{5}^{2}=C_{5}^{3}, C_{4}^{1}=C_{4}^{3}, C_{3}^{1}=C_{3}^{2}$, so the four duplicates should be removed.
$\therefore$ There are $15-5-4=6$ different hyperbolas. Therefore, the answer is (D).