[SOLVED] 33b-counting problem-players and line up

yvonnehr · #1 $Old$ September 5th, 2009, 09:24 PM

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?

yvonnehr · #2 $Old$ September 5th, 2009, 10:23 PM

Quote:

Originally Posted by yvonnehr $View Post$

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.

Rapha · #3 $Old$ September 5th, 2009, 10:26 PM

Hi yvonnehr!

Quote:

Originally Posted by yvonnehr $View Post$

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?

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

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

regards
Rapha

Edit:

Quote:

Originally Posted by yvonnehr $View Post$

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.

Yes, that sounds good.
By the way what is the book's answer?

garymarkhov · #4 $Old$ September 5th, 2009, 11:14 PM

Quote:

Originally Posted by yvonnehr $View Post$

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?

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