2
$\begingroup$

John plans to organize a gift exchange at a party. All the gifts will be put into identical color boxes so no one can tell whose gift is in a box. The boxes will be randomly distributed to the partygoers. It is not desirable for anyone to be assigned his/her own gift. John wants to find out what the probability is that no one is assigned his/her own gift after the exchange.

(a) Suppose there are 4 people attending the party. In how many ways can the boxes be distributed so that exactly one person is assigned his/her own gift?

(b) Suppose instead that there are 5 people attending the party. In how many ways can the boxes be distributed so that no one is assigned his/her own gift? [Hint: You may want to start with a party of only 4 people and consider the complementary event. Then extend your reasoning to the case of 5 people.]

(c) If the boxes are randomly distributed in a party of 5, what is the probability that no one is assigned his/her own gift?

(d) John knows that his best friend Don, one of the other 5 people attending the party, gives exceptionally good gifts. However, John does not feel as confident in his own gift. What is the probability that John ends up receiving Don’ gift, but Don does not receive John’s gift?

I got 8 ways for part a) and 24-15 = 9 ways for part b). Not sure if I'm right. My reasoning was that for part a) if there was an exact match, then the other 3 people can only pick 2 other gifts, so it would be 2x4=8. Same reasoning went with part b), but since it was no match with their own gift, I had the total number of ways minus the ways with an exact match with 5 people. If I'm correct with part b) then part c) should be 9/24? I'm not sure how to start with part d).

  • 2
    For the first two, you are looking for [derangements](https://en.wikipedia.org/wiki/Derangement) – 2017-01-20
  • 0
    Fun fact, adding onto @RossMillikan's link: The ratio of the total number of permutations of $n$ distinct numbered items to the number of permutations of those same items such that no item is in a position matching its index number tends toward $e$ as $n$ approaches infinity. (I [worked this out](http://fora.xkcd.com/viewtopic.php?f=17&t=111107&p=3737460) before having heard of derangements.) :) – 2017-01-20
  • 0
    Am I correct for part a) and b) then? – 2017-01-20
  • 0
    Part (a) is right. Part (b), the answer is $9$ for a party of four - not for a party of five. And the total number of options you were referring to in part (c) was also for a party of four. – 2017-01-20
  • 0
    oh, so then part b) would be 120-15= 105 and part c) would be 105/120? – 2017-01-20
  • 0
    No, the $120$ is right but not $105$. – 2017-01-20
  • 0
    part b) should be 44 and c) would b 44/120? I did part a) by just listing out the possible ways. How would be able to do it using combinations/permutations? Also is there a way to calculate part b) without using derangements? Also, any ideas for part d? @Joffan – 2017-01-23

0 Answers 0