Suppose that 4 guests check their hats when they arrive at a restaurant, and that these hats are returned to them in a random order when they leave. Determine the probability that no guest will receive the proper hat.
My attempt:
So I know that there are 4! ways of "arranging" the hats.
The probability of each receiving their hat is 1/4 and the probability that they will not receive the right hat is 3/4.
I think the answer should be something like:
$\frac{}{4!}$
where the 4! is from the total ways of arranging the hats, and the numerator is the number of "correct" ways to arrange the hats (so no one gets their own hat)But I can't seem to think of how to express that mathematically.
How should I think about this problem instead?
So the book gives the answer 3/8 and myclassmate said the answer is
$\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}=\frac{3}{8}$
But I don't understand why this is?
Best Answer
Use inclusion-exclusion theorem/formula/principle. Total number of ways to distribute 4 hats amon 4 people is $4!$. Now subtract the number of ways 1 person leaves with his hat, add two people left with their hats, subtract three people left with their hats, add 4 ppl left with their hats (the last one is obviously 1).