# the probability of bricks arranged randomly.

• Aug 16th 2012, 08:04 AM
Mhmh96
the probability of bricks arranged randomly.
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?
• Aug 16th 2012, 10:15 AM
Soroban
Re: the probability of bricks arranged randomly.
Hello, Mhmh96!

Quote:

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: .$\displaystyle P(\text{Reds non-adjacent}) \:=\:\frac{70}{495} \:=\:\frac{14}{99}$