35$\cdot$31 A box contains red, white, blue, and green glass balls. In each round, four balls are drawn without replacement, and the following situations are equally likely to occur: (a) four red balls are drawn; (b) one white ball and three red balls are drawn; (c) one white ball, one blue ball, and two red balls are drawn; (d) one ball of each color is drawn. The minimum number of balls that satisfies these conditions is
(A) 19.
(B) 21.
(C) 46.
(D) 69.
(E) more than 69.
(47th American High School Mathematics Examination, 1996)
[Solution] Let the number of red, white, blue, and green balls be $x, y, z, t$ respectively. According to the problem, we have
$$
C_{x}^{4}=C_{x}^{3} C_{y}^{1}=C_{x}^{2} C_{y}^{1} C_{z}^{1}=C_{x}^{1} C_{y}^{1} C_{z}^{1} C_{t}^{1}.
$$

That is, $\frac{x(x-1)(x-2)(x-3)}{24}=\frac{x(x-1)(x-2) y}{6}$
$$
=\frac{x(x-1) y z}{2}=x y z t.
$$

Thus, $y=\frac{x-3}{4}, z=\frac{x-2}{3}, t=\frac{x-1}{2}$.
$\therefore \quad x \equiv 3(\bmod 4) \equiv 2(\bmod 3) \equiv 1(\bmod 2)$
Then $x \equiv-1(\bmod 4) \equiv-1(\bmod 3)=-1(\bmod 2)$
$\therefore x \equiv-1(\bmod 12)$, so $x \geqslant 11$.
Considering (*) equation, thus $y \geqslant 2, z \geqslant 3, t \geqslant 5$.
Therefore,
$$
x+y+z+t \geqslant 21.
$$

Hence, the answer is $(B)$.