27 Then leaving positions i1, . . . ,is alone, we can permute the remaining

n — 8 positions any way we want, for a total of (n — s)! permutations

that leave positions i1, . . . ,is alone. Therefore, the probability that

persons 2'1, . . . ,2", all retrieve their own hats is (n — s)!/'n.!. That is, mm m.— — g—Wl' However these are not quite the probabilities lthat we' re asking about

(though we ’ll want to use them eventually, so remember them), be-

cause if we take .3 = l" and consider the above event that persons

31,. . . ,1", all retrieved their own hat, this event doesn’t preclude the

possiblity that other persons (or even everyone) also retrieved their

own hat. What we want are the probabilities of events like the following: El (31v- win): Ail n“ n Air ﬂ Air“ 0 n A1“: where (11,... , in) Is some permutation of ( , . . . ,n). Event E(i1,---,in)

says that persons i1, . . . ,i, retrieved their own hat but that persons

3,1,1,“ . ,1}, did not. For a particular (i1, . . . ,z'n), that would be one way for the event of 7' persons retrieving their own hat to occur. These

events Emmm are the right events to be considering, because as we

let (1'1, . . . , in) vary over all possible permutations, we get all the possi-

ble ways for exactly 1" persons to retrieve their own hat. However, here

we need to be careful in our counting, because as we vary (:1, . . . ,in)

over all possible permutations we are doing some multiple counting of

the same event. For example, suppose n = 5 and 7' = 3. Then, if you

go examine the way we’ve defined the event E(,-,,,-21,-33,-41,5) in general,

you’ll see that the event E(1,2,3,4,5) is the same as the event E(3,2,1,4’5)

or the event E(3’251’5’4). In general, if we have a particular permutation (i1, . . . , in) and con- sider the event 152mm,”), we can permute the first '1" positions any way