# 1996 AHSME Problems/Problem 26

## Problem

An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:

(a) the selection of four red marbles;

(b) the selection of one white and three red marbles;

(c) the selection of one white, one blue, and two red marbles; and

(d) the selection of one marble of each color.

What is the smallest number of marbles satisfying the given condition? $\text{(A)}\ 19\qquad\text{(B)}\ 21\qquad\text{(C)}\ 46\qquad\text{(D)}\ 69\qquad\text{(E)}\ \text{more than 69}$

## Solution

Let the bag contain $n$ marbles total, with $r, w, b, g$ representing the number of red, white, blue, and green marbles, respectively. Note that $r + w + b + g = n$.

The number of ways to select four red marbles out of the set of marbles without replacement is: $$\binom{r}{4} = \frac{r!}{24\cdot (r -4)!}$$

The number of ways to select one white and three red marbles is: $$\binom{w}{1}\binom{r}{3} = \frac{w\cdot r!}{6\cdot (r - 3)!}$$

The number of ways to select one white, one blue, and two red marbles is: $$\binom{w}{1}\binom{b}{1} \binom{r}{2} = \frac{wb\cdot r!}{2(r-2)!}$$

The number of ways to select one marble of each colors is: $$\binom{w}{1}\binom{b}{1} \binom{g}{1}\binom{r}{1} = wbgr$$ Setting the first and second statements equal, we find: $$\frac{r!}{24\cdot (r -4)!} = \frac{w\cdot r!}{6\cdot (r - 3)!}$$ $$r - 3 = 4w$$

Setting the first and third statements equal, we find: $$\frac{r!}{24\cdot (r -4)!} = \frac{wb\cdot r!}{2(r-2)!}$$ $$(r-3)(r-2) = 12wb$$

Setting the last two statements equal, we find: $$\frac{wb\cdot r!}{2(r-2)!} = wbgr$$ $$r - 1 = 2g$$

These are all the "linking equations" that are needed; the transitive property of equality makes the other three equalities unnecessary.

From the first equation, we know that $r$ must be $3$ more than a multiple of $4$, or that $r \equiv 3 \mod 4$

Putting the first equation into the second equation, we find $r-2 = 3b$. Therefore, $r \equiv 2 \mod 3$. Using the Chinese Remainder Theorem, we find that $r \equiv 11 \mod 12$.

The third equation gives no new restrictions on $r$; it is already odd by the first equation.

Thus, the minimal positive value of $r$ is $11$. This requires $g=\frac{r - 1}{2} = 5$ by the third equation, and $w = \frac{r-3}{4} = 2$ by the first equation. Finally, the second equation gives $b = \frac{(r-3)(r-2)}{12w} = 3$.

The minimal total number of marbles is $11 + 5 + 2 + 3 = \boxed{21}$, which is option $\boxed{B}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 