1996 AHSME Problems/Problem 26

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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 $\boxed{B}$.

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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