There are 16 distinct pairs of gloves in a drawer There are 16 distinct pairs of gloves in a drawer (so, 32 gloves in total).  5 gloves arepicked at random.  What is the probability that among these 5 gloves there is at least one matching pair?Hint: it is easier to compute probability of the complement of this event
I am struggling to solve this...
so I think the denominator is (32 choose 5)
but the top is a bit difficult...
so I need to compute the chance of picking 5 and having NO pairs
my initial thought is (16 choose 5) but I have decided that does not really make sense...
another thought would be using (32*30*28*26*24) / (32*31*30*29*28) but this might be overcounting because of order? but if both have the denominator of 5! to account for removing the overcounting because of order... = 0.8226424422 so 1 - 0.8226424422 ?
thanks for reading
 A: Naturally, it is easier to calculate the complement, because the statement "at least one matching pair" means there could be up to five matching pairs, and rather than computing five separate cases we are better off separating the one remaining case i.e. where there are no matching pairs.
Let us call the gloves as  $(L_1,...,L_{16}) , (R_1,...,R_{16})$ where $L_i , R_i$ are a pair of gloves.
Then, a choice of five gloves, none of which matches its pair, consists of two independent choices :


*

*A choice of five numbers between $1$ and $16$. This can be done in $\binom{16}{5}$ ways.

*For each chosen number, we must assign whether we choose the left of that pair, or the right of that pair. This can be done in $2^5 = 32$ ways, since for each number we have two choices, and there are five chosen numbers.
Now, you can see how this works : for example, suppose I choose $\{1,2,3,4,5\}$ and $(L,L,R,R,R)$ , then we have chosen $1_L,2_L,3_R,4_R,5_R$ : no matching pairs.

The opposite is also true : given five gloves with no matching pairs, they correspond to five different numbers, and of each number either an $L$ or $R$ has been chosen.
Consequently, the answer to the numerator is $\binom{16}5 \times 2^5$. The denominator is correct : it's the number of all possible choices, which is $\binom{32}5$.

Alternately
The alternate thinking is also correct, but needs a little refinement.
Another way to choose a five subset of non-matching gloves is to pick one at a time. So you first have $32$ choices, then $30$ choices, then $28,26,24$ choices at the end. We must divide by $5!$ since we are picking a "set" of gloves i.e. the order of picking does not matter.
So the alternate answer is $\frac{32 \times 30 \times 28 \times 26 \times 24}{5!}$. This is the same as the earlier answer, because we may remove a $2$ from each of the five terms of the numerator, and then multiply top and bottom by $11!$ :
$$
\frac{32 \times 30 \times 28 \times 26 \times 24}{5!} = \frac{2^5 \times 16 \times 15 \times 14 \times 13 \times 12}{5!} = \frac{2^5 \times 16!}{5!11!} = \binom{16}5 \times 2^5
$$
A: Compute the probability of the event when among the $5$ gloves there is no matching pair.
Now you want to choose 5 gloves such that there is no matching so choose from the 16 gloves 
Total number$=\binom{16}{5}\cdot 2^5$
total ways of choosing 5 gloves $\binom{32}{5}$
$$\binom{32}{5}-\binom{16}{5}\cdot 2^5$$
