Question

How many different 9 player teams can be fielded if there are 3 players who can only play catcher, 4 player who can only play first base , 6 player who can only pitch ,and 14 players who can play in any in any of the remain 6 position?

Answer 1

\( \binom{3}{1} \times \binom{4}{1} \times \binom{6}{1} \times \binom{14}{6} = \cdots \)

Answer 2

not match the key

Answer 3

what?

Answer 4

I solve your equation not match with the key is 216,216

Answer 5

answer is 216,216 ?

Answer 6

yes

Answer 7

Yes, that's correct.

\[ \binom{3}{1} \times \binom{4}{1} \times \binom{6}{1} \times \binom{14}{6} = 216216 \]

Answer 8

3C1*4C1*6C1*14C6 ?

Answer 9

Yes.

Answer 10

3* 4* 6* 3003 = 216,216
OK IT correct thank

Answer 11

glad to help :)