Let me see.
Okay lets choose the people one at a time SO order does count.
The number of ways that the 23 people can all have different birthdays is 365C23 I think
The total number of ways 23 people can have birthdays = 365^23
The chance of finding 2 or more with the same birthday would be 1-(365C23)/(365^23)
$${\mathtt{1}}{\mathtt{\,-\,}}{\frac{{\left({\frac{{\mathtt{365}}{!}}{({\mathtt{365}}{\mathtt{\,-\,}}{\mathtt{23}}){!}}}\right)}}{{{\mathtt{365}}}^{{\mathtt{23}}}}} = {\mathtt{0.507\: \!297\: \!234\: \!323\: \!985\: \!4}}$$
which is slightly more than a 50% chance.
Now to my mathematician colleagues. I think that this is correct but can this be done using combinations instead of Permutations?
I canot think of a way
$${{\left({{\mathtt{9}}}^{{\mathtt{9}}}\right)}}^{{\mathtt{9}}} = {\mathtt{196\,627\,050\,475\,552\,913\,618\,075\,908\,526\,912\,116\,283\,103\,450\,944\,214\,766\,927\,315\,415\,537\,966\,391\,196\,809}}$$
A good answer. Thanks Coolkille98
4-6i /1-i +-4+10i / -7+3i
Where do you want the brackets to go?
This is how it should be interpreted when there are no brackets.
$$\\4-(6i /1)-i +-4+(10i / -7)+3i\\\\ 4-6i-4+\frac{-10i}{7}+3i\\\\ -3i+\frac{-10i}{7}\\\\ \frac{-21i}{7}+\frac{-10i}{7}\\\\ \frac{-31i}{7}\\\\$$
No Cooltj
$$10^8$$ only has 8 zeros
this is a very good answer.
Thanks Cooltj
Yes but Sydney is 3am the next day!
+118732 
