For 30:

>! $(!4 - 4) \times \left( \frac{4!}{4} \right)$  
>!   
>! $ =(9 - 4) \times \left( \frac{24}{4} \right) = 5 \times 6 = 30$

For 31:

>! $44 - 4 - !4$  
>!  
>! $= 44 - 4 - 9 = 31$

Note that:

>! $!n$ is the subfactorial of $n$.  
>! For a non-negative-integer $n$ this is the number of **[derangements](https://en.wikipedia.org/wiki/Derangement)** of $n$  
>! (the number of ways to arrange $n$ items such that *no* item is at its naturally ordered position)  
>! This is  
>! $n! \sum_{i=0}^n \frac{(-1)^i}{i!}$  
>!  
>! As such  
>! $!4 = 4! \sum_{i=0}^4 \frac{(-1)^i}{i!} = 24 \times \left(\frac{(-1)^0}{0!} + \frac{(-1)^1}{1!} + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!}\right)$  
>! $= 24 \times \left(\frac{1}{1} + \frac{-1}{1} + \frac{1}{2} + \frac{-1}{6} + \frac{1}{24}\right)$  
>! $= \left(24 - 24 + 12 - 4 + 1\right)$  
>! $= 9$  
>!  
>! Or, using ABCD the 9 derangements are:  
>! 1. BADC  
>! 2. BCDA  
>! 3. BDAC  
>! 4. CADB  
>! 5. CDAB  
>! 6. CDBA  
>! 7. DABC  
>! 8. DCAB  
>! 9. DCBA  
>!  
>! But not any of the other 15 permutations:  
>! ABCD . ACDB . BACD . CABD . DACB  
>! ABDC . ADBC . BCAD . CBAD . DBAC   
>! ACBD . ADCB . BDCA . CBDA . DBCA