A homework question states:

A room holds two rows of six seats each. Two friends are assigned randomly to the 12 seats. What is the probability that the 2 friends sit in adjacent seats?

Note: Friends sitting behind friends don't count. Friends sitting diagonally adjacent to each other don't count. Only friends setting beside each other (left/right) in the same row count.

$$\cdot~~~~~= Empty~seat$$ $$\times = Occupied~seat$$

$$\begin{bmatrix} \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \end{bmatrix}$$

By drawing out the favorable possibilities:

$\begin{bmatrix} \times & \times & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \end{bmatrix} $$\begin{bmatrix} \cdot & \times & \times & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \end{bmatrix}$$ \begin{bmatrix} \cdot & \cdot & \times & \times & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \end{bmatrix} $$\begin{bmatrix} \cdot & \cdot & \cdot & \times & \times & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \end{bmatrix}$$ \begin{bmatrix} \cdot & \cdot & \cdot & \cdot & \times & \times \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \end{bmatrix}$

$\begin{bmatrix} \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \times & \times & \cdot & \cdot & \cdot & \cdot \\ \end{bmatrix} $$\begin{bmatrix} \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \times & \times & \cdot & \cdot & \cdot \\ \end{bmatrix}$$ \begin{bmatrix} \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \times & \times & \cdot & \cdot \\ \end{bmatrix} $$\begin{bmatrix} \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \times & \times & \cdot \\ \end{bmatrix}$$ \begin{bmatrix} \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \times & \times \\ \end{bmatrix}$

It seems like there are a total of 10 favorable situations.

I hope I'm right in saying there are a total of ${12 \choose 2}$ total possible situations (friends can sit in any two seats)?

So is the probability that 2 friends sit adjacent to each other in this room of 12 seats: $$\frac{10}{{12 \choose 2}} = \frac{10}{66} = 0.1515152$$

Whether that's right or wrong, I guess, what's the better mathematical approach (using the whole ${X \choose Y}$ thing to think about this problem?

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The calculation is correct, and efficiently done. There are $\binom{12}{2}$ equally likely ways to select $2$ seats from $12$.
The probability we put Alicia in an end seat is $\frac{4}{12}$, Given she is in an end seat, the probability Bob ends up beside her is $\frac{1}{11}$.
The probability Alicia is not in an end seat is $\frac{8}{12}$, and then Bob has probability $\frac{2}{11}$ of ending up beside her. Thus our probability is $$\frac{4}{12}\cdot\frac{1}{11}+\frac{8}{12}\cdot\frac{2}{11}.$$