can some one just give a hint?

To solve the problem of finding the number of ways the numbers 1 through 5 can be arranged into the five boxes so that the inequalities $\_ < \_ > \_ < \_ > \_$ hold, we need to carefully consider the pattern and constraints.

 

Let's break it down step by step:

 

1. **Pattern Analysis:**

   - The sequence alternates between increasing and decreasing.

   - If we let the positions be labeled as $a < b > c < d > e$, we need to ensure that each set of inequalities is satisfied.

 

2. **Key Points:**

   - The first number must be less than the second number.

   - The second number must be greater than the third number.

   - The third number must be less than the fourth number.

   - The fourth number must be greater than the fifth number.

 

3. **Counting Valid Sequences:**

   - There are a total of 5! (120) possible permutations of the numbers 1 through 5.

   - We need to determine how many of these permutations satisfy the given inequalities.

 

One way to approach this problem is to use a systematic method to generate valid sequences, but a more efficient way is to realize that this problem is a known combinatorial problem.

 

### Generating Function Approach (Optional):

 

This problem can also be solved using generating functions and other combinatorial techniques, but for simplicity, let's use a direct combinatorial argument.

 

### Direct Counting:

 

Instead of listing all permutations, we can use a combinatorial argument based on symmetry and known results in permutations and inequalities.

For a permutation of $n$ elements satisfying a specific pattern of inequalities like this, it can be shown that the number of valid permutations is given by:

 

$$\frac{(n-1)!}{(k-1)! \cdot (n-k-1)!}$$

 

where $k$ is the number of ascents (increasing steps) in the permutation.

 

For the pattern $_ < _ > _ < _ > _$, there are 2 ascents and 2 descents in the sequence.

 

Therefore, the number of valid permutations is:

 

$$\frac{4!}{1! \cdot 1! \cdot 2!} = \frac{24}{2} = 12$$

 

Thus, there are 12 valid ways to arrange the numbers 1 through 5 into the five boxes to satisfy the given inequalities.