John plans to organize a gift exchange at a party. All the gifts will be put int
ID: 3201616 • Letter: J
Question
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. 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? 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? If the boxes are randomly distributed in a party of 5, what is the probability that no one is assigned his/her own gift? John knows that his best friend Galerius, 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 Galerius' gift, but Galerius does not receive John's gift?Explanation / Answer
(A)
Consider there are four boxes W,X,Y and Z. and the correct order of these boxes is WXYZ.
Now, there is case where the W will be assigned correctly and other 3 will not. This can be done in two ways i.e. WYZX and WZXY
Similarly, if we fix X there are 2 combinations possible YXZW and ZXWY
This means there are total 4*2 = 8 ways
(B)
In case of 4 people there are total 4! possible combinations i.e. 24
Consider the case where at least 1 person gets his gift. In case of 4 people this can be done as below
1 person gets his gift, this can be achieved in 8 ways (refer part A)
2 persons get their gift, this can be achieved in 4C2*1 = 6 ways
3 persons get their gift, this can be achieved in 1 way
Total number of ways such that at least 1 person gets his own gift is 8 + 6 + 1 = 15 ways
Number of ways no person will get his own gift = 24 - 15 = 9 ways
In case of 5 people
1 person gets his own gift, can be done in 5C1 * 15 = 75 ways
2 persons get their own gift, can be done in 5C2 * 2 = 10 * 2 = 20 ways
3 persons get their own gift, can be done in 5C3 * 1 = 10 ways
4 persons get their own gift, can be done in 1 way
total number of ways such that at least 1 person gets his own gift = 75 + 20 + 10 + 1 = 106
Hence number of ways that boxes be distributed so that no one is assigned his/her own gift = 120 - 106 = 14 ways
(C)
Probability that no one is assigned his/her own gift = 14/120 = 0.1167
(D)
In this condition, if John recieves Galerius' gift other 4 can be arranged in 4! = 24 ways
but Galerius does not receive John's gift this mean possible ways are 4! - 3! = 24 - 6 = 18
Probability of this event = 18 / 120 = 0.15
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