Combination Challenge – 2025/03/05

Solution: Since $2$ players do not want to be benched, they must be included in the starting lineup. Therefore, the other $4$ players in the starting line-up must be selected from the remaining $6$ players, including those $2$ players who do not want to be captain, vice-captain, or goalie.

Case 1: If none of the two players who do not want to be captain, vice-captain, or goalie is selected.

The total number of ways to form the starting lineup would be $$6!=720$$ as no restrictions will be applied to the positions.

Case 2:If we select one of two players who do not want to be captain, vice-captain, or goalie to be in the starting lineup.

There are $2$ ways to select one of the above two to be one of the $3$ unique field players.

Additionally, $3$ additional players will be selected from the remaining 4 players who do not have any restrictions. These $3$ players plus the other $2$ players who do not want to be on the bench will have $5!$ permutations for the starting lineup.

Therefore, the total number of ways is $$ {{2}\choose{1}}\cdot {{3}\choose{1}}\cdot{{4}\choose{3}}\cdot 5!=2\cdot 3\cdot 4\cdot 120=2880$$

Case 3: If we select both two players who do not want to be captain, vice-captain, or goalie to be in the starting lineup.

There are $3$ ways to select $2$ positions from the $3$ unique field players for the above two players with $2!$ permutations.

The remaining $2$ players will be selected from the $4$ players without restriction. These $2$ players plus the $2$ players who do not want to be on the bench will have $4!$ permutations for the starting lineup.

Therefore the total number of ways is: $${{3}\choose{2}}\cdot 2! \cdot {{4}\choose{2}}\cdot 4!=3\cdot 2\cdot 6\cdot 24=864$$

Combining the above $3$ cases, the answer to the question is $$720+2880+864=\boxed{4464}$$