Baseball Probability With Permutation and Combinations. I am a (Gifted) sixth grader, who after taking the government issued MAP test, got selected test questions based on my weaknesses. One of the test concepts I got was probability, but it was Geometry and used Permutations and Combinations. I know the specific equations concerning permutations and combinations, but this question stumped me. Can anyone help me answer this question?
The baseball coach is going to pick 9 players for the team. He has 9 outfielders, 10 infielders, and 4 pitchers to pick from. The team must have 3 outfielders, 1 pitcher, and 5 infielders. How many ways can he select his baseball team?
I will leave this open-ended, even though there are answer choices, to know all of y'all's answers.
 A: I will assume that the exact positions each player plays is irrelevant beyond the distinction between "outfield / infield / pitcher" and the order of players is irrelevant.  If you wish to specify the batting order or the exact positions played (e.g. first base vs second base vs shortstop, etc...) then you can expand on this.
Apply rule of product and what you know about binomial coefficients.


*

*From the 9 available outfielders, pick which $3$ outfielders are used.

*From the 10 available infielders, pick which $5$ are used.

*From the 4 available pitchers, pick which $1$ is used.
Multiply the total number of ways in which you can complete each step to get the total number.

 There are $\binom{9}{3}$, read aloud as "9 choose 3", ways to pick the outfielders.  This is a binomial coefficient, link above, in case you aren't familiar with them and need to read more.

$~$

 $\binom{9}{3}\cdot \binom{10}{5}\cdot \binom{4}{1}$

A: He needs three outfielders from nine, so he can choose them in ${9 \choose 3}=\frac {9!}{3!(9-3)!}$ ways.  Do the same for each other position and use the multiplication principle.
