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Six digits from $\{2, 3, 4, 5, 6, 7, 8 \}$ are chosen and arranged in a row without replacement. Find the probability the number is divisible by $2$

Total possibilities: $8!/2$

Our event possibilities: $2*7!$

Thus $P = \frac{2*7!}{8!/2} = 1/2$

But this is incorrect according to the textbook. What is wrong?

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    Why so complicated? Divisibility by $2$ depends only on the last digit being even, and there are four even digits out of seven in the set given. – 2017-01-12

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The OP's solution is incorrect, at both stages.

There are seven digits, of which four are even and three are odd. The last digit must be one of those seven, and all seven are equally likely. So the probability that it the six-digit number is even is the probability that the last digit is even, which is $\frac{4}{7}$.

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    I see, I incorrectly counted the set of numbers =) – 2017-01-12