Shrek is love. Shrek is life

Here're some restrictions for this problem.

• $S = 0$ .
• $L, E$ can be any number leftover from the rest.
• $I = O + 1$ and $V + I > 10$ to get residues from $V+I$ for 3rd digit.
• $V \neq 9$ because $\overline{O9} + I = \overline{(O+1)O}$ from above statement.
• $O \neq 9$ because residues from $V+I$ changes $L$

Let's start counting! (sorry for brute force here)

• $O = 2$ gives $I = 3$ and $V = 8$ (1 combination)
• $O = 3$ gives $I = 4$ and $V = 7, 8$ (2 combinations)
• $O = 4$ gives $I = 5$ and $V = 6,7,8$ (3 combinations)
• $O = 5$ gives $I = 6$ and $V = 7,8$ (2 combinations)
• $O = 6$ gives $I = 7$ and $V = 4,5,8$ (3 combinations)
• $O = 7$ gives $I = 8$ and $V = 3,4,5,6$ (4 combinations)
• $O = 8$ gives $I = 9$ and $V = 2,3,4,5,6,7$ (6 combinations)

Total $= 21$ combinations.

So we have 5 numbers left for $L$ and $E$ because $O,I,V,F,S$ used it.

• Choose $2$ numbers from $5$ numbers, $\displaystyle \dbinom{5}{2}$
• $L, E$ can switch order, $2!$ .

Therefore, the number of ways $\displaystyle = 21\times \dbinom{5}{2} \times 2! = \boxed{420}$ combinations!!!

"Shrek is love. Shrek is life."

One of the deepest motto of the world that Adele could roll in it.

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#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{int l,o,v,e,i,s,f,conta=0,love[10000][2],is[10000][2],t=0,exit[10000][2]; for(int h=0;h<10001;h++) for(int z=0;z<3;z++) exit[h][z]=0;
for(int l=1;l<10;l++)
   for(int o=0;o<10;o++)
      for(int v=0;v<10;v++)
         for(int e=0;e<10;e++)
            for(int i=1;i<10;i++)
               for(int s=0;s<10;s++)
                  for(int f=0;f<10;f++)
                     if(l*1000+o*100+v*10+e+i*10+s==l*1000+i*100+f*10+e 
                     && l!=o && l!=v && l!=e && l!=i && l!=s && l!=f && 
                     o!=v && o!=e && o!=i && o!=s && o!=l && o!=f &&
                     v!=e && v!=i && v!=s && v!=l && v!=f &&
                     e!=i && e!=s && e!=l && e!=f && 
                     i!=s && i!=l && i!=f && i!=e &&
                     s!=l && s!=i && s!=f && s!=e &&
                     l!=i && l!=f && l!=e &&
                     i!=f && i!=e &&
                     f!=e) 
                      conta++;
cout<<conta<<"\n";
 system("pause");
return 0;
}

conta=420

Sorry for the bash, here is the easiest and really untidy program, an obvious one, but that gives us the answer...

One sure thing, that we conclude $s=0$ and thus, all other variables are non-zero.

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for l in range(1,10):
    for o in range(1,10):
        for v in range(1,10):
            for e in range(1,10):
                for i in range(1,10):
                    for f in range(1,10):
                            if s ==0 and l!= o and l!=v and l!=e and l!=i and l!=f and o!=v and o!=e and o!=i and  o!=f and v!=e and v!=i and and v!=f and e!=i and e!=f and i!=f and  (1000*l+100*o+10*v+e+10*i+s==1000*l+100*i+10*f+e) :
                                li.append((l,o,v,e,i,s,f)) 

Then we ask for $\textbf{len(li)}$ and the answer is $\boxed{420}$