aime26-19

Proposed solution

I need to find the values of n where the probability of drawing exactly 4 red marbles equals the probability of drawing exactly 5 red marbles when drawing 7 marbles without replacement.

Let me denote:
- $r$ = number of red marbles
- $b$ = number of blue marbles
- $n = r + b$

The probability of drawing exactly $k$ red marbles out of 7 draws is:
$$P(\text{exactly } k \text{ red}) = \frac{\binom{r}{k}\binom{b}{7-k}}{\binom{n}{7}}$$

Setting the probabilities equal:
$$\frac{\binom{r}{4}\binom{b}{3}}{\binom{n}{7}} = \frac{\binom{r}{5}\binom{b}{2}}{\binom{n}{7}}$$

Since $\binom{n}{7}$ appears in both denominators, I can cancel it:
$$\binom{r}{4}\binom{b}{3} = \binom{r}{5}\binom{b}{2}$$

Expanding the binomial coefficients:
$$\frac{r!}{4!(r-4)!} \cdot \frac{b!}{3!(b-3)!} = \frac{r!}{5!(r-5)!} \cdot \frac{b!}{2!(b-2)!}$$

Simplifying:
$$\frac{r(r-1)(r-2)(r-3)}{4!} \cdot \frac{b(b-1)(b-2)}{3!} = \frac{r(r-1)(r-2)(r-3)(r-4)}{5!} \cdot \frac{b(b-1)}{2!}$$

$$\frac{r(r-1)(r-2)(r-3)}{24} \cdot \frac{b(b-1)(b-2)}{6} = \frac{r(r-1)(r-2)(r-3)(r-4)}{120} \cdot \frac{b(b-1)}{2}$$

$$\frac{r(r-1)(r-2)(r-3)b(b-1)(b-2)}{144} = \frac{r(r-1)(r-2)(r-3)(r-4)b(b-1)}{120}$$

Canceling common factors $r(r-1)(r-2)(r-3)b(b-1)$:
$$\frac{b-2}{144} = \frac{r-4}{120}$$

Cross-multiplying:
$$120(b-2) = 144(r-4)$$
$$120b - 240 = 144r - 576$$
$$120b = 144r - 336$$
$$5b = 6r - 14$$
$$b = \frac{6r - 14}{5}$$

Since $b$ must be a positive integer, $6r - 14$ must be divisible by 5.
This means $6r \equiv 14 \pmod{5}$, so $r \equiv 4 \pmod{5}$.

Let $r = 5k + 4$ for integer $k \geq 0$.

Then: $b = \frac{6(5k + 4) - 14}{5} = \frac{30k + 24 - 14}{5} = \frac{30k + 10}{5} = 6k + 2$

So $n = r + b = (5k + 4) + (6k + 2) = 11k + 6$

The constraint is that there are at least 7 marbles of each color:
- $r \geq 7$: $5k + 4 \geq 7$, so $k \geq \frac{3}{5}$, which means $k \geq 1$
- $b \geq 7$: $6k + 2 \geq 7$, so $k \geq \frac{5}{6}$, which means $k \geq 1$

Also, we need $r \leq n - 7$ and $b \leq n - 7$ (since we're drawing 7 marbles).

Since $n = 11k + 6$, we have:
- $r = 5k + 4 \leq 11k + 6 - 7 = 11k - 1$, so $5k + 4 \leq 11k - 1$, which gives $5 \leq 6k$, so $k \geq \frac{5}{6}$, meaning $k \geq 1$
- $b = 6k + 2 \leq 11k + 6 - 7 = 11k - 1$, so $6k + 2 \leq 11k - 1$, which gives $3 \leq 5k$, so $k \geq \frac{3}{5}$, meaning $k \geq 1$

Therefore, $k \geq 1$.

The values of $n$ are: $n = 11k + 6$ for $k = 1, 2, 3, 4, 5, ...$

For $k = 1$: $n = 17$
For $k = 2$: $n = 28$  
For $k = 3$: $n = 39$
For $k = 4$: $n = 50$
For $k = 5$: $n = 61$

Let me verify the first case: $n = 17$, $r = 9$, $b = 8$
- $\binom{9}{4}\binom{8}{3} = 126 \times 56 = 7056$
- $\binom{9}{5}\binom{8}{2} = 126 \times 28 = 3528$

Wait, this doesn't match. Let me recheck my algebra.

Going back to: $5b = 6r - 14$, so $b = \frac{6r - 14}{5}$

Actually, let me be more careful with the constraint that we need at least 7 of each color:
- $r \geq 7$ and $b \geq 7$
- Also, since we draw 7 marbles, we need $r \geq 7$ and $b \geq 7$

Let me recalculate with the constraint that $n = r + b$ where both $r, b \geq 7$.

From $5b = 6r - 14$, we get $b = \frac{6r - 14}{5}$.

For this to be an integer: $6r \equiv 14 \pmod{5}$, so $r \equiv 4 \pmod{5}$.

So $r = 5k + 4$ for some integer $k \geq 0$.
Then $b = \frac{6(5k + 4) - 14}{5} = \frac{30k + 24 - 14}{5} = \frac{30k + 10}{5} = 6k + 2$.

So $n = r + b = (5k + 4) + (6k + 2) = 11k + 6$.

We need $r \geq 7$ and $b \geq 7$.
- $r = 5k + 4 \geq 7 \Rightarrow k \geq \frac{3}{5}$, so $k \geq 1$
- $b = 6k + 2 \geq 7 \Rightarrow k \geq \frac{5}{6}$, so $k \geq 1$

Let me check $k = 1$: $r = 9$, $b = 8$, $n = 17$
$\binom{9}{4}\binom{8}{3} = 126 \times 56 = 7056$
$\binom{9}{5}\binom{8}{2} = 126 \times 28 = 3528$

This is wrong. Let me recheck the constraint.

Actually, I should verify that $k = 1$ gives valid values:
$r = 9$, $b = 8$, so $n = 17$
Check: $\binom{9}{4}\binom{8}{3} = 126 \times 56 = 7056$
$\binom{9}{5}\binom{8}{2} = 126 \times 28 = 3528$

These are not equal. Let me rech