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In a kickball competition of 9 teams, how many possible ma
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19 Oct 2013, 05:46
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In a kickball competition of 9 teams, how many possible matches can each team play with each other?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
------------
Did you try this? How about +KUDOS for me?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
------------
Did you try this? How about +KUDOS for me?
Manager
Joined: 18 Dec 2012
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Re: In a kickball competition of 9 teams, how many possible ma
[#permalink]
19 Oct 2013, 06:00
19 Oct 2013, 06:00
amgelcer wrote:
In a kickball competition of 9 teams, how many possible matches can each team play with each other?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
------------
Did you try this? How about +KUDOS for me?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
------------
Did you try this? How about +KUDOS for me?
Hi amgelcer.
Each team can play with each other in 9C2 ways
9C2 = 9*8 / 1*2 = 36
Option D
Hope it helped
Cheers
Qoofi
Re: In a kickball competition of 9 teams, how many possible ma
[#permalink]
19 Oct 2013, 07:58
19 Oct 2013, 07:58
Expert Reply
Qoofi wrote:
amgelcer wrote:
In a kickball competition of 9 teams, how many possible matches can each team play with each other?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
------------
Did you try this? How about +KUDOS for me?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
------------
Did you try this? How about +KUDOS for me?
Hi amgelcer.
Each team can play with each other in 9C2 ways
9C2 = 9*8 / 1*2 = 36
Option D
Hope it helped
Cheers
Qoofi
Qoofi is correct. In case you need a general reminder of the combination formula as well, here it is:
nCr, read "n choose r" is n!/r!(n-r)!
Here, then we have 9 teams chosen 2 at a time. So 9C2 is 9!/2!(9-2)! --> 9!/2!7! --> 9*8/2.
That may help on future combination problems. Cheers!
Re: In a kickball competition of 9 teams, how many possible ma
[#permalink]
20 Oct 2013, 12:25
20 Oct 2013, 12:25
Expert Reply
amgelcer wrote:
In a kickball competition of 9 teams, how many possible matches can each team play with each other?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
------------
Did you try this? How about +KUDOS for me?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
------------
Did you try this? How about +KUDOS for me?
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Hope it helps.
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Re: In a kickball competition of 9 teams, how many possible ma
[#permalink]
08 Sep 2019, 08:06
08 Sep 2019, 08:06
Expert Reply
Top Contributor
amgelcer wrote:
In a kickball competition of 9 teams, how many possible matches can each team play with each other?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
A DIFFERENT approach
If each of the 9 teams plays every team, then each team plays 8 games (since a team can't play itself)
So, the total number of games = (9)(8) = 72
From here we need to recognize that each game has been COUNTED TWICE.
For example, if Team A and Team B play a game, then Team A counts it as a game, and Team B ALSO counts it as a game.
So, to account for the DUPLICATION, we'll divide 72 by 2 to get 36
Answer: D
Cheers,
Brent
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Location: United States (CA)
Re: In a kickball competition of 9 teams, how many possible ma
[#permalink]
12 Sep 2019, 10:27
12 Sep 2019, 10:27
Expert Reply
amgelcer wrote:
In a kickball competition of 9 teams, how many possible matches can each team play with each other?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
------------
Did you try this? How about +KUDOS for me?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
------------
Did you try this? How about +KUDOS for me?
The total possible number of matches is 9C2 = 9! / (2! x 7!) = (9 x 8) / 2 = 36.
Answer: D
_________________
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Scott@TargetTestPrep.com
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In a kickball competition of 9 teams, how many possible ma
[#permalink]
07 Oct 2019, 19:44
07 Oct 2019, 19:44
amgelcer wrote:
In a kickball competition of 9 teams, how many possible matches can each team play with each other?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
9+8+7+.....+1 = (8*9)/2 = 36
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10372
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: In a kickball competition of 9 teams, how many possible ma
[#permalink]
20 Feb 2021, 07:43
20 Feb 2021, 07:43
Expert Reply
Total teams: n
Number of matches: \(\frac{n(n-1) }{ 2}\)
=> n = 9. Hence, number of matches: \(\frac{9 * 8 }{ 2}\) = 36
Answer D
Number of matches: \(\frac{n(n-1) }{ 2}\)
=> n = 9. Hence, number of matches: \(\frac{9 * 8 }{ 2}\) = 36
Answer D
gmatclubot
Re: In a kickball competition of 9 teams, how many possible ma [#permalink]
20 Feb 2021, 07:43
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