Suppose that we have 10 different pairs of gloves. From the 20 gloves, 4 are chosen at random. What is the probability of getting at least one pair?

Please note that my solution differs from the authors’ one given at the end

There are a total of \(\binom{20}{4} = 4845\) possible outcomes. There are \(10 \cdot \binom{18}{2} = 1530\) outcomes with at least one pair and \(\binom{10}{2} = 45\) outcomes with exactly two pairs, which were already counted in the first equation so need to be subtracted to get the final number of \(10 \cdot \binom{18}{2} - \binom{10}{2} = 1485\) outcomes. The final probability is thus \(\frac{10 \cdot \binom{18}{2} - \binom{10}{2}}{\binom{20}{4}} = \frac{1485}{4845} = \frac{99}{323}\).

Alternatively, we can first count the number of outcomes that do not have a single pair in them. There are \(\frac{20 \cdot 18 \cdot 16 \cdot 14}{4!} = 3360\) of them, so we need to subtract them from the total and divide by the total to get the same final probability of \(\frac{\binom{20}{4} - \frac{20 \cdot 18 \cdot 16 \cdot 14}{4!}}{\binom{20}{4}} = \frac{1485}{4845}\)

Authors’ answer:

\[\frac{\frac{20 \cdot 18 \cdot 16 \cdot 14}{4!}}{\binom{20}{4}}\]¹