Answer to Question #152269 in Discrete Mathematics for Inchara br

Question

There are 7 groups in a picnic who has brought their own lunch box, and then the 7 lunch box are exchanged within those groups. Determine the number of ways that they can exchange the lunch box such that none of them can get their own.

Answer 1

Let there are $n$ people's and each people's have its own assigned boxes.

Then total number of derangement so that none of them get their own box is

$=n![1-\frac {1}{1!}+\frac {1}{2!}-\frac {1}{3!}+\frac {1}{4!}-.....(-1)^n.\frac {1}{n!}]$

According to the problem, there are 7 groups in picnic and each group has its own lunch box.

There the total number of ways that they can exchange the lunch box such that none of them can get their own is

$=7![1-\frac {1}{1!}+\frac {1}{2!}-\frac {1}{3!}+\frac {1}{4!}-\frac {1}{5!}+\frac {1}{6!}-\frac {1}{7!}]$

$=[\frac{7!}{2!}-\frac{7!}{3!}+\frac{7!}{4!}-\frac{7!}{5!}+\frac{7!}{6!}-\frac{7!}{7!}]$

$=[2520-840+210-42+7-1]$

$=1854$