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?
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?
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.
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?
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