Probability that the cards are in AP 
Box $1$ contains three cards bearing numbers $1, 2, 3$ ; box $2$ contains five cards bearing numbers $1, 2, 3, 4, 5$ ; and box $3$ contains seven cards bearing numbers $1, 2, 3, 4, 5, 6, 7$. A card is drawn from each of the boxes. Let $x_i$ be the number on the card drawn from the $i$th box, $i = 1, 2, 3.$ What's the  probability that $x_1 , x_2 , x_3$ are in an arithmetic progression?

My attempt: Total cases are: $3\cdot5\cdot7=105$. Favorable instances are $10$.
$ \{ (1,1,1), (1,2,3), (1,3,5), (1,4,7), (2,2,2), (2,3,4), (2,4,6), (3,3,3), (3,4,5), (3,5,7)\}$.
So, my answer is $\frac{10}{105}$, but the answer is given as $\frac{11}{105}$. Which case am I missing?

As others have commented, $(3,2,1)$ is missing. Instead of listing the choices, observe that $x_1,x_2,x_3$ are in AP iff $x_1+x_3=2x_2$. This implies either both $x_1,x_3$ are odd or both are even since $x_1+x_3$ is even. We have $2$ odd values of $x_1$ and $4$ odd values for $x_3$, giving $2\times 4$ pairs. We also have $1$ even value of $x_1$ and $3$ even values of $x_3$, giving $1\times 3$ pairs.
For all pairs $(1+1)/2=1\le x_2=(x_1+x_3)/2\le(3+7)/2=5$ i.e. $x_2\in\{1,2,...,5\}$. Total number of pairs is $8+3=11$.
