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?
Best Answer
The calculation is correct, and efficiently done. There are $\binom{12}{2}$ equally likely ways to select $2$ seats from $12$.
You then counted the "favourables" well, though it was quite unnecessary to list separately the favourables in the first row and the favourables in the second.
Since you did the problem in a nice way, let me do it an uglier way. Call the people Alicia and Bob, and seat Alicia first. We could (i) put her in an end seat or (ii) not in an end seat.
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 $$\frac{4}{12}\cdot\frac{1}{11}+\frac{8}{12}\cdot\frac{2}{11}.$$