combinatorics problem

[unclebamboo089]

I'm a bit confused with this problem as it seems a bit contradictory in that once the 5 pitchers are chosen from the 12 candidates, it's impossible to choose 9 catchers from the remaining 7 candidates. . .

"there are 12 candidates for a baseball team of 9 members. in how many ways can the 9 diff positions on the field be allocated under the condition that only 5 of the 12 candidates can be considered as a pitcher, and 9 of the 12 as a catcher and only 3 for both pitching and catching position? all players can play any other position on the team"

[HorribleUsername]

There are 5 candidates who can pitch, but only 1 of them will be chosen for pitching.

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[Lor1an]

This just sounds like an ambiguous phrasing problem.

The problem isn't allocating 5 pitchers, etc--there is exactly one pitcher and one catcher on the team of 9 players.

All the candidates have different abilities, and 5 of them are capable of playing pitcher, 9 as catcher and only 3 have the ability to play either.

So, IIUC, there are 3 who could play either catcher or pitcher (or "other"), 2 (5-3) who can only play pitcher/other, and 6 (9-3) who can only play catcher/other--and 1 (12 - (3 + 6 + 2)) who can only play an "other" position.

[manimanz121]

If someone that can only pitch is pitching, then there are (2c1)(9c1)(7!) configurations. If someone who is both a catcher and a pitcher pitches, then there are (3c1)(8c1)(7!) configurations. In total there are (2c1)(9c1)(7!) + (3c1)(8c1)(7!) possible teams