Combinatorics Problem: To order or not to order. The question I am working on is:
A Little League team that has 15 players on
its roster.
a.How many ways are there to select 9 players for the
starting lineup?
b.How many ways are there to select 9 players for the
starting lineup and a batting order for the 9 starters?
c. Suppose 5 of the 15 players are left-handed. How many
ways are there to select 3 left-handed outfielders and have
all 6 other positions occupied by right-handed players?
First of all, what specifically is a starting line-up? and would the order be pertinent or impertinent? How about for batting order? Obviously the name sort of implies that order matters; however, I don't want to presume anything.
 A: They are apparently treating a lineup as an ordered list of the nine starting players. This is unusual; perhaps they’re thinking that the list starts out as a list of the $9$ playing positions (catcher, pitcher, first base, etc.) which then gets filled in with some set of $9$ names; in that case listing Joe as pitcher would be different from listing Joe as catcher. Thus, they arrive at $\binom{15}99!=\frac{15!}{6!}$ ways to fill out the lineup card.
They then treat the batting order as a distinct ordering of the $9$ starting players, so each of the $\binom{15}99!$ lineups can bat in any of $9!$ orders, and we get an answer of $\binom{15}9(9!)^2$.
For the last question, there are $\binom53$ ways to choose $3$ left-handed players and $\binom{10}6$ ways to choose $6$ right-handed players; that’s $2100$ ways to choose the people. There are then $3!$ ways to permute the lefties amongst their $3$ assigned positions and $6!$ ways to permute the righties amongst theirs, for a total of $2100\cdot3!\cdot6!=9,072,000$ arrangements.
And this is a very badly worded problem!
