<s>[INST] 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.
 [/INST] 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
$$ g \cdot d^2 \cdot f^3  \cdot \frac { G^3 }{3!}\frac { D^7 }{7!}\frac { F^{10} }{10!}\frac { V^4 }{4!}$$
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