I don't understand how to do this or which formulas to use.

A little league team has 15 players.

How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters?

Printable View

- Sep 5th 2009, 08:24 PMyvonnehr[SOLVED] 33b-counting problem-players and line up
I don't understand how to do this or which formulas to use.

A little league team has 15 players.

How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters? - Sep 5th 2009, 09:23 PMyvonnehrpossible solution. please check.
Well, I think the book's answer may be wrong on this one. Here is my solution. Is this correct?

(15 choose 9) x 9! = 181681894400

"15 choose 9" represents the number of ways you can pick 9 players from the 15 on the roster, and 9! represents the possible ordered lineups. I multiplied them together because because we are matching one set of outcomes to the other set of outcomes. - Sep 5th 2009, 09:26 PMRapha
Hi yvonnehr!

To select 9 players out of 15 there are $\displaystyle \begin{pmatrix} 15\\ 9 \end{pmatrix}$ different ways.

After that there are 9 players and nine different positions (1,2,3,4...,8,9).

There are 9! ways tochoose the batting order.

So the solution of this problem is

$\displaystyle \begin{pmatrix} 15 \\ 9 \end{pmatrix}*9!$

regards

Rapha

Edit:

Yes, that sounds good.

By the way what is the book's answer? - Sep 5th 2009, 10:14 PMgarymarkhov
Are there two questions here? If you start by selecting 9 players for the starting lineup and don't care about their batting order, there are 15!/(6!*9!) = 5005 combinations. Then, once you start caring about batting order, there are 15!/6! = 1816214400 permutations.

By the way, I used to get confused by combinations and permutations all the time. No more after reading Easy Permutations and Combinations | BetterExplained