Previously on the birthday problem .........................
i found out that nPr = n!/(n-r)!
can I use this, well I have 2 objects, eric and noddy, and 4 days to select their birthdays, so I suppose those would be the distinct objects, although distinct to me means different and I want to find out the same days.Anyway let's try it, n =4 and r = 2

nPr = 4 !/(4-2)! =4*3*2*1/2*1 =24/2 =12

nPr = 4 !/(4-2)! =4*3*2*1/2*1 =24/2 =12

12 ? This being the number of ways they do not share, so distinct it is ? , but where are my 16 possibilities. Returning to probabilites let me send noddy home and take a bag of four smarties of different colours. Picking one , returning it is like my birthday thing but sweeter. what is the chance that i will pick two of the same colour in four tries. If the first is yellow 1/4 , no no, no this is all too confusing at the moment need a smoke break.