相邻偶数

先排列奇数 $1,3,5,7$，共有 $4!=24$ 种方法。

奇数排好后，会形成 5 个可插入偶数的位置：

_ O _ O _ O _ O _

其中 $O$ 表示奇数，空位表示可放偶数的位置。现在要把 3 个偶数放进这 5 个空位里，而且每个空位最多放 1 个偶数，因此选空位的方法有

$${5\choose 3}=10$$

3 个偶数自身还可以排列 $3!=6$ 种方法，所以总数为

$$4! \times {5\choose 3} \times 3! = \boxed{1440}$$

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Original Explanation

We start by determining how many ways we can arrange the odd digits (1, 3, 5, 7). There are $4! = 24$ ways to arrange them.

This creates a gap where event digits can be placed.

_ O _ O _ O _ O _

Here O are the odd digits, _ are gaps where event digits can go. The number of gaps = the number of odd digits + 1 = 5.

We have 3 even digits to place into these 5 gaps. Since no two even digits can be in the same gap, we have ${5}\choose{3}$ $=10$ ways. Arranging the even digits amongst themselves gives $3!=6$.

Putting it together

$$4! \times {5\choose 3} \times 3! = \boxed{1440}$$