Question

A certain class has 20 students, and meets on Mondays and Wednesdays in a classroom with exactly 20 seats. In a certain week, everyone in the class attend both days. On both days, the students choose their seats completely randomly (with one student per seat). Find the probability that no one sits in the same seat on both days of that week.

Answer 1

The probability that no one sits in the same seat on both days of that week is given by,
`P(\cap^(20)_(i=1)A_i^c)=(1)/(e)`

Explanation:

Given : A certain class has 20 students, and meets on Mondays and Wednesdays in a classroom with exactly 20 seats. In a certain week, everyone in the class attend both days. On both days, the students choose their seats completely randomly (with one student per seat).

To find : The probability that no one sits in the same seat on both days of that week ?

Solution :

Let
`A_i` be the i-th student sits on seat which he has been sitting on Monday.

According to question,

We have to calculate
`P(\cap^(20)_(i=1)A_i^c)`

Applying inclusion exclusion formula,

`P(\cap^(20)_(i=1)A_i^c)=1-P(\cap^(4)_(i=1)A_i)`

`P(\cap^(20)_(i=1)A_i^c)=1-P(A_1)+...+P(A_(20))-P(A_1\cap A_2)+...+P(A_(19)\cap A_(20))+P(A_1\cap A_2\cap A_3)+...+P(A_(18)\cap A_(19)\cap A_(20))....-P(A_1\cap A_2...\cap A_(20))`

Using symmetry,

`P(\cap^(20)_(i=1)A_i^c)=1-\sum^(20)_(k=1)(-1)^(k+1)\binom{20}{k}P(A_1\cap ...\cap A_k)`

`P(\cap^(20)_(i=1)A_i^c)=1-\sum^(20)_(k=1)(-1)^(k+1)\binom{20}{k}((20-k)!)/(20!)`

`P(\cap^(20)_(i=1)A_i^c)=1+\sum^(20)_(k=1)(-1)^(k)(1)/(k!)`

`P(\cap^(20)_(i=1)A_i^c)=\sum^(20)_(k=0)(-1)^(k)(1)/(k!)`

`P(\cap^(20)_(i=1)A_i^c)=(1)/(e)`

Therefore, The probability that no one sits in the same seat on both days of that week is given by,
`P(\cap^(20)_(i=1)A_i^c)=(1)/(e)`