# 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.

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From context it’s clear that in this problem the starting lineup is just the unordered set of $9$ players who will start the game; imposing a batting order makes it an ordered set of $9$ players. – Brian M. Scott Feb 2 '13 at 21:18
I'm sorry, but I don't see how it is clear. – Mack Feb 2 '13 at 21:30
The only way that starting lineup could reasonably be interpreted to imply a specific order would be if that order were the batting order; there’s no other ordering that would make any sense. And it’s clear from the wording of (b) that the batting order is not determined by the starting lineup, so the starting lineup must refer just to the set of players. – Brian M. Scott Feb 2 '13 at 21:34
Here one has to know the culture of baseball to work on a math question! – Maesumi Feb 2 '13 at 22:07
@Maesumi: Could be worse. There are $20$ cricketers in the village of Little Piddle in the Marsh. Four are fast bowlers. Six others have soft hands and exceptional reflexes. Only one is willing to keep wicket. In how many ways can the captain field two fast bowlers, three slips, and a gully? – Brian M. Scott Feb 2 '13 at 23:07

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!

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When you say, "...the list starts out as a list of the 9 playing positions (catcher, pitcher, first base, etc.)," does that correspond to ${{15}\choose{9}}$, which says out of the 15 positions how many ways can you arrange 9 of them? – Mack Feb 2 '13 at 22:43
@Eli: He starts out with a list of the nine playing positions. He chooses $9$ of his $15$ players; that can be done in $\binom{15}9$ ways. And then he assigns each of the $9$ chosen players to a position and writes his name down opposite that position; that can be done in $9!$ ways. Thus, the whole list can be filled out in $\binom{15}99!$ different ways. – Brian M. Scott Feb 2 '13 at 22:47
I'm sorry, I honestly can't see why you perform a combination and then multiply by $9!$. For a), I did a permutation and I got the same answer. For b), I did multiplied $P_{15,9} \cdot P_{15,9}$, but was off from the answer by a factor of 2 and I can't figure out why. – Mack Feb 3 '13 at 16:13
@EliMackenzie: Suppose that there were just $3$ positions, called Left, Right, and Centre, and he had $5$ players, A, B, C, D, and E. He could choose $3$ of the $5$ in $\binom53=10$ ways, so he could field $10$ different teams of $3$. Say he chooses A, C, and D. Now he has to specify who plays which position. There are $3!$ different orders in which he can list them, so there are $3!$ ways that he can specify who plays Left, who plays Right, and who plays Centre. That is, that same team of $3$ can fill the $3$ positions in $3!$ different ways. Total: $\binom533!$ lineups. – Brian M. Scott Feb 3 '13 at 20:24