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> 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?
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**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?