How many possible cases are there

Question

I'm a dentist with biiiig gaps in my math. People have normally 32 teeth. If we say any random tooth or teeth can be extracted, what are the different possible cases (is it called permutations?) Are there? Someone can have all 32 teeth present someone else can have the 1molar extracted. Some one third can have 2nd molar extracted... Until all teeth are extracted. Is it factorial of 32? Thank you and excuse my lack of math knowledge

Answer 1

Number the positions of all 32 teeth, then you can describe any case as a string of 32 zeros or ones , with the nth character being 1 if the tooth is present and 0 if the tooth has been extracted.

The number of such strings is $2^{32} \approx 4.3 \times 10^9$ this number includes every possible case.

The number $32! \approx 2.6 \times 10^{35}$ would be the number of patterns you could make by painting each tooth one of 32 different colours with no two teeth having the same colour. Also, it is the number of ways that you could rearrange all 32 teeth.