Why is the probability for the signal at 160 in the bromine mass spectrum twice as high as for the signals at 158 and 162? The mass spectrum in Bromine, with the molecules $\ce{^{158}Br2+}$, $\ce{^{160}Br2+}$ and $\ce{^{162}Br2+}$:


As you can see, the $\ce{^{160}Br2+}$ is almost double in intensity compared to the $\ce{^{158}Br2+}$ and the $\ce{^{162}Br2+}$ peak. 

The book I am reading simply states that this is because 


  The probability of two different isotopes occurring in a $\ce{Br2}$ molecule are twice that of the same isotope appearing in a $\ce{Br2}$ molecule.


This is supported by the $\ce{^{160}Br2+}$ peak, formed from the $\ce{^{79}Br}$ and $\ce{^{81}Br}$ isotopes. 
Likewise, $\ce{^{158}Br2+}$ peak is formed from two $\ce{^{79}Br}$ isotopes and $\ce{^{162}Br2+}$ is formed from two $\ce{^{81}Br}$ isotopes. 

However, I am confused by the explanation given by the book above. 
Why is the probability of two different isotopes occurring in a $\ce{Br2}$ molecule twice that of the same isotope appearing in a $\ce{Br2}$ molecule?
All possible arrangements of $\ce{Br2}$ molecule:


$\displaystyle 79 + 79 = 158$
$\displaystyle \color{red}{79 + 81} = 160$
$\displaystyle \color{red}{81 + 79} = 160$
$\displaystyle 81 + 81 = 162$


The amount of $\ce{^{79}Br}$ and $\ce{^{81}Br}$ in nature is roughly the same, thus each permutation is equally probable. There are two arrangements that lead to $160$. While $158$ and $162$ each have only one arrangement. Therefore $160$ is twice as likely to be found compared to other masses.