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.
 A: Hint
Suppose defence didn't choose any from the versatile pool of $4$, # of  ways would be
$\quad\quad\quad\quad\Large[\binom31\binom72][\binom{10}3+\binom{10}2\binom41+\binom{10}1\binom42+\binom43]$
Complete for cases where defence chooses 1 or 2 from the versatile pool
A: Consider cases depending on the number of players who can play either position are used.
Suppose exactly $k$ players who can play either position are used, where $0 \leq k \leq 4$.  If $d$ of these $k$ players are selected to play defense, where $0 \leq d \leq 2$, then the remaining $k - d = f$ of these players must play forward, where $0 \leq f \leq 3$.  If $d$ of the players who could play two positions play defense, then the remaining $2 - d$ defensive players must be selected from the $7$ players who only play defense.  If $f$ of the players who could play two positions play forward, then the remaining $3 - f$ forwards must be selected from $10$ players who only play forward.  The goalie must be selected from one of the three players who only play goalie.  Hence, if $k$ players who could play either defense or forward are selected to play on the team and $d$ of these $k$ players are selected to play defense, then the number of ways to form such a team is
$$\binom{3}{1}\binom{4}{k}\binom{k}{d}\binom{7}{2 - d}\binom{10}{3 - f}$$
I will leave it to you to finish the calculations.
A: 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
$$ g \cdot d^2 \cdot f^3  \cdot \frac { G^3 }{3!}\frac { D^7 }{7!}\frac { F^{10} }{10!}\frac { V^4 }{4!}$$
