# Probabilty of an event of combination problem

I have eight positions $P_1,P_2,P_3,P_4,P_5,P_6,P_7,P_8$ to be filled with eight numbers $x_1,x_2,x_3,x_4$ and $y_1,y_2,y_3,y_4$ with following rules:

1. $P_1,P_2,P_3,P_4$ should be filled with two of $x_i$ and two of $y_i$ in which if one $x_i$ in $P_1$ other $x_i$ in $P_3$ or if one $x_i$ in $P_2$ other $x_i$ in $P_4$.

2. $P_5,P_6,P_7,P_8$ should be filled with remaining two $x_i$ and two $y_i$ in which if one $x_i$ in $P_5$ other $x_i$ in $P_7$ or if one $x_i$ in $P_6$ other $x_i$ in $P_8$.

Order is also matters.

Can someone help me to calculate the probability of this event?

Multiply the following:

• Number of ways to choose $2$ out of $4$ x's: $\binom42=6$
• Number of ways to choose $2$ out of $4$ y's: $\binom42=6$
• Number of ways to place the 1st x: $4$
• Number of ways to place the 1st y: $2$
• Number of ways to place the 2nd x: $1$
• Number of ways to place the 2nd y: $1$
• Number of ways to place the 3rd x: $4$
• Number of ways to place the 3rd y: $2$
• Number of ways to place the 4th x: $1$
• Number of ways to place the 4th y: $1$

Hence the total number of ways is $6\cdot6\cdot4\cdot2\cdot1\cdot1\cdot4\cdot2\cdot1\cdot1=2304$.

• So probability is $2304/8!$? – Frey Nov 14 '16 at 7:43
• @Frey: You're welcome :) – barak manos Nov 14 '16 at 7:45