My Math Forum Permutation/Combination

 July 8th, 2010, 06:04 PM #1 Newbie   Joined: Jul 2010 Posts: 2 Thanks: 0 Permutation/Combination A certain baseball team has 24 players. Only nine can be on the field at a time. Each of the nine players on the field has a distinct field position: pitcher, catcher, first baseman, second baseman, third baseman, short stop, left field, right field, or center field. Assume for the moment that every player is qualified to play every position. How may ways are there to select nine of the 24 players to be on the field (without regard to which position each player will have)? My reasoning is: 24x23x22x21x20x19x18x17x16 =474467051520 I don't get why this is wrong though. Any help would be appreciated!
 July 8th, 2010, 06:10 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Permutation/Combination 24 choose 9, which is 24! / (9! * *24-9)!). Your answer counts each team 9! times.
July 9th, 2010, 06:41 AM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Re: Permutation/Combination

Hello, Louis Felipe!

Since it says "without regard to which position each player will have",
[color=beige]. . [/color]the listing of the nine positions was unnecessary and grossly misleading.

Quote:
 A certain baseball team has 24 players. Only nine can be on the field at a time. How may ways are there to select nine of the 24 players to be on the field? My reasoning is: 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 = 474,467,051,520 I don't get why this is wrong.

CRGreathouse is absolutely correct.

We are selecting 9 players from the available 24 players.
[color=beige]. . [/color]This is a combinations problem.

[color=beige]. . [/color]$_{24}C_9 \;=\;{24\choose9} \;=\;\frac{24!}{9!\,15!} \;=\;1,307,504$

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### how may ways are there to select nine of the 21 players to be on the field (without regard to which position each player will have)?

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