# Mock AIME I 2012 Problems/Problem 9

## Problem

A class of seven students is doing a Secret Santa, for which all seven students have contributed a gift each. However, one of the seven students has not arrived yet. The teacher decides to randomly assign each of the six students present to a gift that is not their own. If $P$ is the probability that the seventh student is left with his own gift once he arrives, find $\lfloor1000P\rfloor$.

## Solution

Given that each of the six students receives a gift that is not their own, there are just two disjoint possibilities: one of those six students received the seventh student's gift (and therefore the seventh student will not receive his own gift), or none of those six students received the seventh student's gift (and therefore the seventh student will receive his own gift). In the first case, this is simply the number of ways to rearrange the seven gifts so that no one receives his/her own gift, or $!7$. In the second case, this is simply the number of ways to rearrange the first six students' gifts such that no one receives his/her own gift, or $!6$. Thus, the desired probability is $P=\frac{!6}{!6+!7}$. We note that $$!7 = 7\cdot 6!\left(-\frac{1}{7!}+\sum_{n=0}^{6}\frac{(-1)^n}{n!}\right) = 7 \cdot !6-1.$$ Thus, $P=\frac{!6}{8\cdot!6 - 1}$, which is just larger than $\frac{1}{8}$. By estimating the magnitude of $!6$ (or calculating exactly $!6 =265$), it is quite clear that the difference of $-1$ on the denominator is negligible to the desired calculation. We have then that $\lfloor 1000P \rfloor = 1000\cdot \frac{1}{8} = \boxed{125}$.

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