- #1

- 996

- 5

## Homework Statement

N men throw their hats on to the floor, mix them up, and draw at random. What is the probability that no man gets his own hat

## Homework Equations

$$P(E_1^c E_2^c \dots E_N^c) = 1-P(E_1 \cup E_2 \dots \cup E_N)$$

$$P(\text{n-tuple}) = \frac{1}{n!}$$

## The Attempt at a Solution

The way I've seen this explained appears to me to have an inconsistency that I don't know how to resolve.

1). Total # arrangements of N hats among N men is [itex]N![/itex] This is the denominator in the probability expression.

2). Define [itex]E_i[/itex] as the event that n=1 man chooses his own hat. Imagine man #1 choosing a hat first. There is only 1 way he can choose his own hat. The remaining [itex](N-1)[/itex] hats can be arranged [itex](N-1)![/itex] ways. But, since each man can choose first, we must multiply by [itex]C_1^N = N[/itex]. Thus [itex]P(E_1) = P(E_2) \dots = P_N) = N (N-1)!/N! = 1 = \frac{1}{n!}[/itex]

3) Same argument for k=2, k=3, etc. gives the result stated above.

Here is what doesn't make sense to me. If we argue that there are N ways that a man can choose first, that implies order of choosing is important. If order is important, then the order of choosing 2nd, 3rd, etc. is also important, which isn't accounted for in the above argument.

What am I missing?