if someone have 4 red bricks and 8 blue bricks ,he want to arrange them randomly in circular shape, What is the probability that two red bricks are not side by side?
Hello, Mhmh96!
If someone have 4 red bricks and 8 blue bricks and arranges them randomly in a circle,
what is the probability that two red bricks are not side by side?
There are $\displaystyle \frac{12!}{4!\,8!} \,=\, 495$ possible arrangements.
(This includes duplication from rotations.)
Place the 8 blue bricks in a circle, leaving a space between them.
. . $\displaystyle \begin{array}{cccccccc}&& B & \circ & B \\ & \circ &&&& \circ \\ B &&&&&& B \\ \circ &&&&&& \circ \\ B &&&&&& B \\ & \circ &&&& \circ \\ && B & \circ & B \end{array}$
Select 4 of the 8 spaces and insert the red bricks.
There are: $\displaystyle \frac{8!}{4!\,4!} \,=\,70$ ways.
Therefore: .$\displaystyle P(\text{Reds non-adjacent}) \:=\:\frac{70}{495} \:=\:\frac{14}{99}$