Ice Hockey

## 3911810

https://math.stackexchange.com/questions/3911810

In how many ways can a basketball coach form a team of 2 centers, 2 guards and 1 forward out of a total of 10 basketball players, if 3 players can only play center, 3 only guard position, 1 only forward position, 2 players guard or forward, 1 center or forward?

$144$ or $162$ ? Reviewing the problem in the spirit of GF/egf approach

p
l
a
y

r
e
s
t

1
c
.
C

c
2

C
2

2
!

1
C

C
2

2
!

C
3

3
!

1
g
.
G

g
2

G
2

2
!

1
G

G
2

2
!

G
3

3
!

1
f
.
F

1
F

1
(
g
f
)
.
G
F
(

g
2

g
f

f
2

)

G

F
2

2
!

1
G
F

G

F
2

2
!

1
(
c
f
)
.
C
F

1
C
F

{\displaystyle
\begin{array}{c|c|c}
 play & rest & \\
\hline
 1 + c.C + c^2\frac {C^2}{2!} & 1+C+\frac {C^2}{2!} +\frac {C^3}{3!} \\
 1 + g.G + g^2\frac {G^2}{2!} & 1+G+\frac {G^2}{2!} +\frac {G^3}{3!} \\
 1 + f.F & 1+F \\
 1+ (g+f).GF + (g^2 + gf + f^2)\frac {GF^2}{2!} & 1+GF+\frac {GF^2}{2!}\\
 1+ (c+f).CF & 1+CF \\
\hline
\end{array}
}

Above C, G, F, GF and CF denote the five sorts of players. These variables contributes to the e.f.g part since all players have their own identities.

The small letters c, g, f are counters and they make the G.F. part.

The GF/egf polynomial $play\cdot rest$ is the product of the ten factors in the table.

we are now interested in the coefficient of

c
2

g
2

f

C
3

3
!

G
3

3
!

F
1

1
!

G

F
2

2
!

G

F
2

2
!

{\displaystyle c^2g^2f\frac { C^3 }{3!}\frac { G^3 }{3!}\frac { F^1 }{1!} \frac { GF^2 }{2!} \frac { GF^2 }{2!} }

in the GF/egf polynomial, which is $144$.

However, if instead of $(g^2 + g \cdot f + f^2)$ we take $(g+f)^2$ we get the $162$.
This coefficient expresses the fact that two GF players could fill two distinct positions, or they are just part of the team.

## 4533892

https://math.stackexchange.com/questions/4533892

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?

Given the explanation here
https://math.stackexchange.com/questions/3911810/combination-problem-picking-a-basketball-team-with-restrictions one could expect two solutions, 50916 and 42273.

p
l
a
y

r
e
s
t

1
g
.
G

1
G

G
2

2
!

G
3

3
!

1
d
.
D

d
2

D
2

2
!

1
D

D
2

2
!

⋯

D
7

7
!

1
f
.
F

f
2

F
2

2
!

f
3

F
3

3
!

1
F

F
2

2
!

⋯

F

10

10
!

o
p
t
i
o
n

A
)
,

B
)

1
V
⋯

V
4

4
!

{\displaystyle
\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
$$ g \cdot d^2 \cdot f^3 \cdot \frac { G^3 }{3!}\frac { D^7 }{7!}\frac { F^{10} }{10!}\frac { V^4 }{4!}$$

b

c

d