4 People Gift Exchange

Question

4 people are exchanging gifts. How many combinations are there so that no one receives their own gift? I tried this problem myself, and got 3!. My friends told me that it's 9.

I got 3! because I thought:

Person A: Has 3 options for gifts (excluding his own) Person B: Has 2 options for gifts (excluding his own, and the one taken) Person C: Has 1 option for a gift (excluding his own, the two taken) Person D: Has 1 option for a gift (excluding the 3 taken)

So I deduced that the answer would be 3!

But my friends did it by hand and got 9, one of them showed me this chart:

A B C D

B A D C

B C D A

B D A C

The chart is all the possibilities if A takes gift B, and he told me to repeat it 3 times for all other gifts that A could take→ 3 + 3 + 3 = 9

Is there an elegant solution to this problem? Could someone explain to me why my method neglects to include the other 3 options.

Any help is greatly appreciated.

I read «combinatronics» and I came to see what it was. Color me disappointed! — Mariano Suárez-Álvarez, Feb 08 '15 at 01:57
Sorry, I'm not really sure what it's called. I just saw this problem online and wanted to solve it. — Jon, Feb 08 '15 at 02:04
You are write that A has 3 choices but one choice is gift B and, in that one case, if B goes next she has 3 and not two choices. — Aaron Meyerowitz, Feb 08 '15 at 03:06

score 4 · Accepted Answer · edited Apr 13 '17 at 12:21

4

This is called a derangement of the set of $4$ elements. A derangement is a permutation in which no element is left unmoved. There are three approaches to computing derangements in this answer. One method gives $$ 4!\left(\frac1{0!}-\frac1{1!}+\frac1{2!}-\frac1{3!}+\frac1{4!}\right)=9 $$

edited Apr 13 '17 at 12:21

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answered Feb 08 '15 at 02:16

robjohn

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4 People Gift Exchange

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