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# There are 10 people to play in the tournament in which a

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Manager
Joined: 04 Dec 2005
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There are 10 people to play in the tournament in which a [#permalink]

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22 Jul 2006, 09:37
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"There are 10 people to play in the tournament in which a team of 3 will play another team of 3 in each game. How many different games can be scheduled in the rournament?"

4200
1400
120
35
20
CEO
Joined: 20 Nov 2005
Posts: 2896
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 25

Kudos [?]: 290 [0], given: 0

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23 Jul 2006, 22:02
Ways in which first team can be selected: 10C3 = 120
Ways in which secon dteam can be selected: 7C3 = 35

Total matches possible = (120 * 35)/3 = 1400

We are diving by three to remove duplicates. If there are two members in a team then we will divide by 2.

Is this correct??
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Manager
Joined: 04 Dec 2005
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Kudos [?]: 23 [0], given: 0

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24 Jul 2006, 19:46
Ways in which first team can be selected: 10C3 = 120
Ways in which second team can be selected: 7C3 = 35
Total matches possible = (120 * 35) = 4200
Intern
Joined: 01 Jul 2006
Posts: 19
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24 Jul 2006, 21:05
ps_dahiya wrote:
Ways in which first team can be selected: 10C3 = 120
Ways in which secon dteam can be selected: 7C3 = 35

Total matches possible = (120 * 35)/3 = 1400

We are diving by three to remove duplicates. If there are two members in a team then we will divide by 2.

Is this correct??

Can you explain the division by three and removal of duplicates ?
CEO
Joined: 20 Nov 2005
Posts: 2896
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 25

Kudos [?]: 290 [0], given: 0

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25 Jul 2006, 08:11
bondguy wrote:
ps_dahiya wrote:
Ways in which first team can be selected: 10C3 = 120
Ways in which secon dteam can be selected: 7C3 = 35

Total matches possible = (120 * 35)/3 = 1400

We are diving by three to remove duplicates. If there are two members in a team then we will divide by 2.

Is this correct??

Can you explain the division by three and removal of duplicates ?

Lets take an example of 5 person and two members in a team.

Ways of selecting first team = 5C2 = 10
Ways of selecting second team = 3C2 = 3
All 30 cases are below. Duplicates are marked. Half of them are duplicates. Because there are two people in team thats why only half are unique. If there are three people in a team then there will be only 1/3 unique cases.

12---34
12---35
12---45
13---24
13---25
13---45
14---23
14---25
14---35
15---23
15---24
15---34
23---14----DUPLICATE
23---15----DUPLICATE
23---45
24---13----DUPLICATE
24---15----DUPLICATE
24---35
25---13----DUPLICATE
25---14----DUPLICATE
25---34
34---12----DUPLICATE
34---14----DUPLICATE
34---25----DUPLICATE
35---12----DUPLICATE
35---14----DUPLICATE
35---24----DUPLICATE
45---12----DUPLICATE
45---13----DUPLICATE
45---23----DUPLICATE

Hope this helps.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

25 Jul 2006, 08:11
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