Question

A drawer contains a mixture of red socks and blue socks, at most 17 in all. It so happens that when two socks are selected randomly without replacement, there is a probability of exactly 1/2 that both are red or both are blue. The largest possible number of red socks in the drawer that is consistent with this data is.

Answer 1

10

Answer 2

Let $R$ be the number of red socks and $B$ be the number of blue socks. The total number of socks is $N = R + B$. We are given that $N \le 17$.

The number of ways to select two socks from $N$ is $\binom{N}{2} = \frac{N(N-1)}{2}$.

The number of ways to select two red socks is $\binom{R}{2} = \frac{R(R-1)}{2}$.

The number of ways to select two blue socks is $\binom{B}{2} = \frac{B(B-1)}{2}$.

The probability that both socks are red or both socks are blue is given as $1/2$.

This probability is $P(\text{both red}) + P(\text{both blue}) = \frac{\binom{R}{2}}{\binom{N}{2}} + \frac{\binom{B}{2}}{\binom{N}{2}}$.

$\frac{\frac{R(R-1)}{2} + \frac{B(B-1)}{2}}{\frac{N(N-1)}{2}} = \frac{1}{2}$

$\frac{R(R-1) + B(B-1)}{N(N-1)} = \frac{1}{2}$

$2[R(R-1) + B(B-1)] = N(N-1)$

Substitute $B = N-R$:

$2[R(R-1) + (N-R)(N-R-1)] = N(N-1)$

$2[R^2 - R + (N^2 - NR - N - NR + R^2 + R)] = N^2 - N$

$2[2R^2 - 2NR + N^2 - N] = N^2 - N$

$4R^2 - 4NR + 2N^2 - 2N = N^2 - N$

$4R^2 - 4NR + N^2 - N = 0$

We can solve this quadratic equation for $R$:

$R = \frac{-(-4N) \pm \sqrt{(-4N)^2 - 4(4)(N^2 - N)}}{2(4)} = \frac{4N \pm \sqrt{16N^2 - 16(N^2 - N)}}{8}$

$R = \frac{4N \pm \sqrt{16N^2 - 16N^2 + 16N}}{8} = \frac{4N \pm \sqrt{16N}}{8} = \frac{4N \pm 4\sqrt{N}}{8} = \frac{N \pm \sqrt{N}}{2}$

For $R$ to be an integer, $N$ must be a perfect square, and $N \pm \sqrt{N}$ must be even. If $N=k^2$, then $R = \frac{k^2 \pm k}{2} = \frac{k(k \pm 1)}{2}$. Since $k(k \pm 1)$ is always even, $R$ is an integer if $N$ is a perfect square.

Also, we must have $N \ge 2$ for the selection of two socks to be possible, so $\binom{N}{2} > 0$.

The possible perfect squares for $N \le 17$ are $1, 4, 9, 16$. Since $N \ge 2$, we consider $N \in \{4, 9, 16\}$.

Case 1: $N=4$. $R = \frac{4 \pm \sqrt{4}}{2} = \frac{4 \pm 2}{2}$. $R_1 = \frac{4+2}{2} = 3$. If $R=3$, $B=N-R=4-3=1$. $(R, B) = (3, 1)$. $R_2 = \frac{4-2}{2} = 1$. If $R=1$, $B=N-R=4-1=3$. $(R, B) = (1, 3)$. Both solutions are valid since $R, B \ge 0$. The possible values for $R$ are 1 and 3.

Case 2: $N=9$. $R = \frac{9 \pm \sqrt{9}}{2} = \frac{9 \pm 3}{2}$. $R_1 = \frac{9+3}{2} = 6$. If $R=6$, $B=N-R=9-6=3$. $(R, B) = (6, 3)$. $R_2 = \frac{9-3}{2} = 3$. If $R=3$, $B=N-R=9-3=6$. $(R, B) = (3, 6)$. Both solutions are valid since $R, B \ge 0$. The possible values for $R$ are 3 and 6.

Case 3: $N=16$. $R = \frac{16 \pm \sqrt{16}}{2} = \frac{16 \pm 4}{2}$. $R_1 = \frac{16+4}{2} = 10$. If $R=10$, $B=N-R=16-10=6$. $(R, B) = (10, 6)$. $R_2 = \frac{16-4}{2} = 6$. If $R=6$, $B=N-R=16-6=10$. $(R, B) = (6, 10)$. Both solutions are valid since $R, B \ge 0$. The possible values for $R$ are 6 and 10.

The possible values for the number of red socks $R$ are $\{1, 3, 6, 10\}$.

We are looking for the largest possible number of red socks.

The largest value in the set $\{1, 3, 6, 10\}$ is 10.