In how many ways can the ice hockey team be formed? 
A ice hockey team consists of 1 goaltender, 2 defencemen and 3 forwards. The coach has available to him: 3 goaltenders, 7 defencemen, 10 forwards and 4 players that can play both position of a defenceman and a forward. In how many ways can the team be formed?


If there would not be players that can play two positions the solution would be:
$$\binom{3}{1}\cdot \binom{7}{2}\cdot \binom{10}{3} = 7560$$
ways. So my question is how do I take the 4 players that can play two different positions into account in the calculation? Obviously if the four players has been chosen to be defencemen, they can't be chosen to be forwards.

Given the explanation here
Combination problem - picking a basketball team with restrictions one could expect two solutions, 50916 and 42273.
\begin{array}{c|c}
  play & rest & \\
\hline
 1 + g.G  & 1 + G + \frac {G^2}{2!} + \frac {G^3}{3!}  \\
 1 + d.D + d^2\frac {D^2}{2!}   & 1 + D + \frac {D^2}{2!} + \cdots + \frac {D ^7}{7!}  \\
 1 + f.F +f^2\frac {F ^2}{2!} + f^3\frac {F^3}{3!}               &  1 + F + \frac {F ^2}{2!} + \cdots + \frac {F^{10}}{10!} \\
 option \ A), \ B)   & 1 + V + \cdots +\frac {V^4}{4!}\\ 
\hline
\end{array}
Option $A)$
$$1 + (f+d) \cdot V + (f+d)^2 \frac {V^2}{2!}+(f+d)^3 \frac {V^3}{3!} +(f+d)^4 \frac {V^4}{4!}$$
Option $B)$
$$1 + (f+d) \cdot V + (f^2+fd+d^2) \frac {V^2}{2!}+(f^3+ffd+fdd+ d^3)\frac {V^3}{3!} +(f^4+fffd+ffdd+fddd + d^4) \frac {V^4}{4!}$$
The GF/egf is the product of the eight factors in the table.
we are now interested in the coefficient of
$$ g \cdot d^2 \cdot f^3  \cdot \frac { G^3 }{3!}\frac { D^7 }{7!}\frac { F^{10} }{10!}\frac { V^4 }{4!}$$
