In a closet there are n pairs of shoes. From the closet, 2r…
In a closet there are $n$ pairs of shoes. From the closet, $2r$ shoes are chosen at random ($2r\lt n$). What is the probability that among the chosen shoes:
$1^{\circ}$ there is no complete pair;
$2^{\circ}$ there is exactly one complete pair;
$3^{\circ}$ there are exactly two complete pairs?
Explanation of Method
All sets of $2r$ shoes out of $2n$ are equally likely: $\binom{2n}{2r}$ outcomes. To build a set with exactly $k$ complete pairs, we choose which $k$ pairs are complete, then choose the pairs that contribute one shoe each, and for each of those decide whether it is the left or the right shoe. The factor $2$ for "left or right" is what makes shoes different from ordinary objects.
Step 1: All outcomes
There are $2n$ shoes. The number of ways to choose $2r$ of them:
\[ \binom{2n}{2r}. \]
Step 2: $1^{\circ}$ — no complete pair
The $2r$ shoes must come from $2r$ different pairs:
- choose which $2r$ pairs are used: $\binom{n}{2r}$ ways;
- from each of them take the left or the right shoe: $2^{2r}$ ways.
\[ \boxed{P_0 = \frac{\binom{n}{2r}\,2^{2r}}{\binom{2n}{2r}}.} \]
Step 3: $2^{\circ}$ — exactly one complete pair
- choose the complete pair: $n$ ways (both shoes are taken);
- the other $2r-2$ shoes come from $2r-2$ different pairs among the remaining $n-1$: $\binom{n-1}{2r-2}$ ways;
- from each of these take the left or right shoe: $2^{2r-2}$ ways.
\[ \boxed{P_1 = \frac{n\binom{n-1}{2r-2}\,2^{2r-2}}{\binom{2n}{2r}}.} \]
Step 4: $3^{\circ}$ — exactly two complete pairs
- choose the two complete pairs: $\binom n2$ ways;
- the other $2r-4$ shoes from $2r-4$ different pairs among the remaining $n-2$: $\binom{n-2}{2r-4}\,2^{2r-4}$ ways.
\[ \boxed{P_2 = \frac{\binom n2\binom{n-2}{2r-4}\,2^{2r-4}}{\binom{2n}{2r}}.} \]
Step 5: The general pattern
For exactly $k$ complete pairs ($0\le k\le r$):
\[ P_k = \frac{\binom nk\binom{n-k}{2r-2k}\,2^{2r-2k}}{\binom{2n}{2r}}, \]
and $P_0+P_1+\dots+P_r = 1$.
Step 6: Example — $n=10$ pairs, $2r=4$ shoes
| $k$ pairs | Count | Probability |
|---|---|---|
| $0$ | $\binom{10}{0}\binom{10}{4}2^{4}=3360$ | $0.6935$ |
| $1$ | $\binom{10}{1}\binom{9}{2}2^{2}=1440$ | $0.2972$ |
| $2$ | $\binom{10}{2}\binom{8}{0}2^{0}=45$ | $0.0093$ |
| total | $\binom{20}{4}=4845$ | $1$ ✓ |
Even with only $4$ shoes out of $10$ pairs, the chance of getting at least one complete pair is about $31\%$.
Result
| Part | Probability |
|---|---|
| $1^{\circ}$ no pair | $\binom{n}{2r}2^{2r}\big/\binom{2n}{2r}$ |
| $2^{\circ}$ exactly one pair | $n\binom{n-1}{2r-2}2^{2r-2}\big/\binom{2n}{2r}$ |
| $3^{\circ}$ exactly two pairs | $\binom n2\binom{n-2}{2r-4}2^{2r-4}\big/\binom{2n}{2r}$ |