How many 5 digit numbers with at least two digits prime and two digits square? 
How many 5 digit numbers with at least two digits prime and at least two digits square? 

All digits are different and non-zero.
So I have seen this problem online and can’t find a solution so just wondering (i) if I’m correct and (ii) if there are any other ways out there.
My solution:
At least two digits prime and two digits square means that at most one digit has neither of these properties (N). We have 3 square (S) numbers and 4 prime numbers (P) in the range also.
Hence the cases are
NSSPP -> $2 \times 3 \times 2 \times 4 \times 3 =144$
SSSPP -> $3 \times 2 \times 1 \times 4 \times 3 =72$
SSPPP -> $3 \times 2 \times 4 \times 3 \times 2 =144$
Total = $360$ 
How’s my solution looking?
If it’s wrong why? If correct, are there any other methods
Edit: changes 6 to 2 in the first line as only neither digits are 6 and 8
 A: As you observe we have three cases:

SSPPN: In order to count this typo we do the following:
  
  
*
  
*Choose $2$ digits square from the set of digits square: you can do this in $\binom{3}{2}$ ways. 
  
*Choose $2$ digits prime from the set of digits prime: you can do this in $\binom{4}{2}$ ways.
  
*Choose $1$ digits neither from the set of digits neither: you can do this in $\binom{2}{1}$ ways.
  
*Since the digits are all different, you can permute them as you want: so you have $5!$ ways to permute.
  
  
  So for this case you have:
  \begin{equation}
\binom{3}{2}\cdot \binom{4}{2} \cdot \binom{2}{1} \cdot 5! = 4320
\end{equation}

And

SSSPP: You have:
  
  
*
  
*$3$ digits square from the set of digits square: you can do this in $\binom{3}{3}$ ways. 
  
*$2$ digits prime from the set of digits prime: you can do this in $\binom{4}{2}$ ways.
  
*Since the digits are all different, you can permute them as you want: so you have $5!$ ways to permute.
  
  
  So for this case you have:
  \begin{equation}
\binom{3}{3}\cdot \binom{4}{2} \cdot 5! = 720
\end{equation}

And

SSPPP: In order to count this typo we do the following:
  
  
*
  
*$2$ digits square from the set of digits square: you can do this in $\binom{3}{2}$ ways. 
  
*$3$ digits prime from the set of digits prime: you can do this in $\binom{4}{3}$ ways.
  
*Since the digits are all different, you can permute them as you want: so you have $5!$ ways to permute.
  
  
  So for this case you have:
  \begin{equation}
\binom{3}{2}\cdot \binom{4}{3} \cdot 5! = 1440
\end{equation}

So the result is $4320+720+1440=6480$.
