I didn't understand Birthday Paradox completely >>>> is it that in every 23 person there is a 50% chance to find a two persons have the same

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  

First ask what's the probability that none of them have a birthday in common.  Put the individuals in order from first to 23rd.

The probability that the 2nd person doesn't have the same birthday as the first is 364/365.

The probability that the 3rd doesn't have the same birthday as either of the first two is 363/365.

The probability that the 4th doesn't have the same birthday as any of the first three is 362/365.

...

...

The probability that the 23rd doesn't have the same birthday as any of the others is 343/365.

Therefore the overall probability that no one has a birthday in common is obtained by multiplying all the above probabilities together.

This is 364*363*362*...*343/365²² = 364!/((365-23)!*365²²) = 365!/((365-23)!*365²³)

The probability that at least two people have the same birthday is therefore 1 - 365!/((365-23)!*365²³) as Melody has noted.

Thanks Alan