Math Help - [SOLVED] cinema problem

1. [SOLVED] cinema problem

4 friends go to the cinema and reserve 4 tickets. The seats are already assigned to them. When they arrive to the cinema, they sit at random. The probability of 1 stay in its place is greater or minor than or equal to the probability of 2 sit in its place?

2. Hello, JoanF!

4 friends go to the cinema and reserve 4 tickets.
The seats are already assigned to them.
When they arrive to the cinema, they sit at random.

The probability that one is in the right seat is:
. . (1) greater than
. . (2) less than
. . (3) equal to
the probability that two are in the right seats?
There are $4! = 24$ possible seatings.

Suppose the four friends are: . $\{A,B,C,D\}$
. . and their assigned seats are in this order: . $A\:B\:C\:D$

One in the right seat.

There are 4 choices for the One.
. . Suppose it is $A.$

Then there are 2 ways for the others to sit in the wrong seats:
. . $A\:C\:D\:B\;\;\text{ and }\;\;A\:D\:B\:C$

Hence, there are: . $4 \times 2 \:=\:8$ ways.

. . $P(\text{one in the right seat}) \:=\:\frac{8}{24}\:=\:\frac{1}{3}$

Two in the right seats.

There are: . ${4\choose2} = {\color{blue}6}$ choices for the Two.
. . Suppose they are $A$ and $B.$

Then there is 1 way for the other two to be in the wrong seats:
. . $A\:B\:D\:C$

Hence, there are: . $6 \times 1 \:=\:6$ ways.

. . $P(\text{two in the right seats}) \:=\:\frac{6}{24}\:=\:\frac{1}{4}$

Therefore: . $P(\text{one in right seat}) \;\;{\color{red}>}\;\;P(\text{two in the right seats})$