Probability of pair of gloves selection In his wardrobe, Fred has a total of ten pairs of gloves. He had to pack his suitcase before a business meeting, and he chooses eight gloves without looking at them. We assume that any set of eight gloves has an equal chance of being chosen.
I am told to calculate the likelihood that these 8 gloves do not contain any matching pairs, i.e. that no two (left and right) gloves are from the same pair.
This is what I came up with, that is, the probability of success for each choice:
$$\frac{20}{20}×\frac{18}{19}×\frac{16}{18}×...×\frac{6}{13}=\frac{384}{4199}≈0.09145$$
At first, I was a little confused by the wording but I believe this seems about right.
Is there an alternative way to get the desired probability, e.g. with $1-...$?
Thanks in advance for any feedback.
 A: There is a more general formula for this.
Here you are asked that no pair is selected, but this formula will take care of any number of pairs selected
With $10$ be the  number of pairs, and $k$ the number of pairs selected from $8$ gloves, the formula is
$\dfrac{\dbinom{10}{k}\dbinom{10-k}{8-2k}\cdot2^{8-2k}}{\dbinom{20}{8}}$
For the particular case for $k=0$, it simplifies to
$\dfrac{\dbinom{10}{0}\dbinom{10-0}{8-2\cdot0}\cdot2^{8-2\cdot0}}{\dbinom{20}{8}}$
$= \dfrac{\dbinom{10}{0}\dbinom{10}8
\cdot2^8}{\dbinom{20}{8}}$
A: We can use a combinatoric argument if you like: there are $20 \choose 8$ ways we could possibly choose $8$ gloves from the $20,$ neglecting order.
To see how many of these will involve us choosing no pairs, we can think about first choosing which pairs we will take one glove from, and then from that choosing what glove to pick from each pair. There are $10$ pairs so we have $10 \choose 8$ ways to choose our pairs, and then for each set of pairs there are $2^8$ ways that we can choose to take the left or right glove from each.
So, if all possible sets of gloves are equally likely to be taken, the probability of taking no pairs of gloves should be $$\frac{{10 \choose 8} \cdot 2^8}{20 \choose 8} = \frac{\frac{10!}{2! 8!} \cdot 2^8}{\frac{20!}{12!8!}} = \frac{(10 \cdot 9 \cdot \ldots \cdot 3) \cdot 2^8}{20 \cdot 19 \cdot \ldots \cdot 13} = \frac{20 \cdot 18 \cdot \ldots \cdot 6}{20 \cdot 19 \cdot \ldots \cdot 13}$$
corroborating your result.
A: Another approach is to use inclusion-exclusion.
There are $\binom{20}{8}$ ways to select the $8$ gloves, all of which we assume are equally likely.  Let's say a selection has "Property $i$" if it includes both gloves of pair $i$, for $1 \le i \le 10$, and let $S_j$ be the total probability (with over-counting) of the selections with $j$ of the properties, for $1 \le j \le 4$.  So
$$S_j = \frac{\binom{10}{j} \binom{20-2j}{8-2j}}{\binom{20}{8}}$$
By inclusion-exclusion, the probability of a selection with none of the properties, i.e. with no pair of matching gloves, is
$$1-S_1+S_2-S_3+S_4 = 0.0914503$$
