S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Fri, 22 Aug 2014 18:40:36 UTC

All times are UTC [ DST ]




Post new topic Reply to topic  [ 5 posts ] 
Author Message
PostPosted: Fri, 27 Apr 2012 17:25:51 UTC 
Offline
Member

Joined: Wed, 15 Feb 2012 07:19:33 UTC
Posts: 31
A judo club has 10 members, of whom 2 are heavy weight players, 5 are middle weight players and 3 are light weight players. The club wants to send a team of 7 players to participate in a competition. The team must consist of a heavy weight player, 2 middle weight players, and 2 reserves from any weight category.

Find the number of ways in which the team can be selected.

My answer is simply 2C1 x 5C2 x 7C4 (remaining 4 players inclusive of reserves) = 420, but the answer given is 600. I'm not sure if it's something in the phrasing of the question i missed out, can anyone explain why? thanks!


Top
 Profile  
 
PostPosted: Fri, 27 Apr 2012 18:17:04 UTC 
Offline
Senior Member

Joined: Tue, 16 Aug 2011 23:38:56 UTC
Posts: 124
thesocialnetwork wrote:
A judo club has 10 members, of whom 2 are heavy weight players, 5 are middle weight players and 3 are light weight players. The club wants to send a team of 7 players to participate in a competition. The team must consist of a heavy weight player, 2 middle weight players, and 2 reserves from any weight category.

Find the number of ways in which the team can be selected.

My answer is simply 2C1 x 5C2 x 7C4 (remaining 4 players inclusive of reserves) = 420, but the answer given is 600. I'm not sure if it's something in the phrasing of the question i missed out, can anyone explain why? thanks!


On the above sentence 2 light weight players is missing.

The correct sentence is:
The team must consist of a heavy weight player, 2 middle weight players, 2 light weight players and 2 reserves from any weight category.


Top
 Profile  
 
PostPosted: Sat, 28 Apr 2012 02:26:32 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame

Joined: Mon, 19 May 2003 19:55:19 UTC
Posts: 8111
Location: Lexington, MA
Hello, thesocialnetwork!

You have a typo in the problem . . .
And I don't agree with their answer either.


Quote:
A judo club has 10 members of whom 2 are heavyweights, 5 are middleweights and 3 are lightweights.

The club wants to send a team of 7 players to participate in a competition.
The team must consist of 1 heavyweight, 2 middleweight players, and 4 reserves from any weight category

Find the number of ways in which the team can be selected.

My answer is simply \left(_2C_1\right)\cdot\left(_5C_2\right)\cdot\left(_7C_4) \:=\: 420 . This is incorrect.
. . but the answer given is 600.

. . \begin{array}{ccc}_2C_1 &=& 2 \\ _5C_2 &=& 10 \\ _7C_4 &=& 35 \end{array}


\text{Answer: }\:2\cdot10\cdot35 \:=\:700



Top
 Profile  
 
PostPosted: Sat, 28 Apr 2012 06:09:13 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6815
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
thesocialnetwork wrote:
A judo club has 10 members, of whom 2 are heavy weight players, 5 are middle weight players and 3 are light weight players. The club wants to send a team of 7 players to participate in a competition. The team must consist of a heavy weight player, 2 middle weight players, and 2 reserves from any weight category.

Find the number of ways in which the team can be selected.

My answer is simply 2C1 x 5C2 x 7C4 (remaining 4 players inclusive of reserves) = 420, but the answer given is 600. I'm not sure if it's something in the phrasing of the question i missed out, can anyone explain why? thanks!


Let's go by fam's interpretation.

If you have to name your reserves, then it is
\displaystyle\binom{2}{1}\binom{5}{2}\binom{3}{2}\binom{5}{2}=600
On the other hand, if you don't have to name the reserves, then your team could have (heavy,middle,light) = (1,3,3), (2,2,3),(2,3,2), or (1,4,2). So there are
\displaystyle\binom{2}{1}\binom{5}{3}\binom{3}{3}+\binom{2}{2}\binom{5}{2}\binom{3}{3}+\binom{2}{2}\binom{5}{3}\binom{3}{2}+\binom{2}{1}\binom{5}{4}\binom{3}{2}
=90
ways to name your team.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
PostPosted: Sat, 28 Apr 2012 06:23:18 UTC 
Offline
Member

Joined: Wed, 15 Feb 2012 07:19:33 UTC
Posts: 31
thanks for all the help! it's appreciated.


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 5 posts ] 

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA