A problematic way of thinking on arranging 7 boys and 3 girls in a row The question is the same as the one asked here 7 boys and 3 girls
how many ways to arrange them in a row so that both ends are boys and no girls adjacent.
I clearly understand the two approaches to solve this permutation problem.
First Approach, arrange the 7 boys first (7!)
                next, there are 6 available slots for 3 girls, that's P(6,3)
    so total is 7! * P(6,3)
Second Approach, Arrange the 3 girls first (3!)
   next, first assume there is no difference between 7 boys
     1. arrange 4 boys into the 4 available slots ( 1 way to arrange)
     2. distribute the 3 rest boys arbitrarily into the  4 slots, that's
        (3+4-1,3)
     3. now consider the full permutation of the boys 7!
      so the answer is 3! * (6,3) * 7! = 7! * P(6,3), which agrees with the first answer
Now the trouble comes when I tried to think in the below way to achieve the same permutation and I failed to spot what went wrong .
  Third (problematic) approach
    1. first arrange the 3 girls, that's 3!
    2. then arrange 4 boys into the 4 slots, that's P(7,4)
    3. now there are 3 boys left
       the first one will have 8 available slots to be placed ( 8)
       the second one will have 9 available slots to be placed (9)
       the last one will have 10 available slots to be placed (10)
    so the answer is 3! * P(7,4) * 8 *9 * 10. 
    Obviously this doesn't agree with the answer and obviously it is wrong.
could you help me to spot the reasoning flaw(s) in the last approach?

The issue is that you are overcounting the solutions with your final approach.  Suppose we have a couple of boys called Adam and Bob, and a couple of girls called Carol and Diana.
Now suppose we consider a permutation that looks something like: ... Carol Adam Bob Diana ... .
In your counting scheme, we would count this multiple times - first when Adam is chosen as one of the first four boys (going between Carol and Diana) and then Bob is chosen (to go between Adam and Diana), and second when Bob is chosen first (to go between Carol and Diana) and then Adam (to go between Carol and Bob).
The number of ways each permutation is overcounted depends on how the boys are distributed between the girls, so it is not a constant.  That makes it difficult to get to a clean expression using this approach.
