1. ## Basic Question

I need to know if I'm doing this problem accordingly

A drawer contains n=5 different and distinguishable pairs of socks (total of 10). If a person randomly selects four socks what is the probability that there is no matching pair in the sample?

solution:

$1-[(4)(\frac{4!}{10!}) - (4)(\frac{1}{10!}) + \frac{1}{10!}]$

2. Hello, lllll

Using $10!$, you seem to be working with permutations.

A drawer contains 5 different and distinguishable pairs of socks (total of 10).
If a person randomly selects four socks, what is the probability that
there is no matching pair in the sample?
There are: . ${10\choose4} \:=\:{\color{blue}210}$ possible samples.

Suppose the pairs of socks are of five different colors.
To have no matching pairs, we must have four socks of different colors.

There are: . ${5\choose4} \:=\:5$ choices for the four colors.
Then we must select one sock from each of those pairs.
. . There are: . ${2\choose1}{2\choose1}{2\choose1}{2\choose1} \:=\:16$ ways.

Hence, there are: . $5\cdot16 \:=\:{\color{blue}80}$ ways to get four socks of different colors.

Therefore: . $P(\text{no matching pairs}) \;=\;\frac{80}{21} \;=\;\boxed{\frac{8}{21}}$

3. why would it not be a permutation, doesn't the order matter?