Problem Solving

Source: Veritas Prep

There are 16 teams in a soccer league, and each team plays each of the others once. Given that each game is played by two teams, how many total games will be played?

A) 256
B) 230
C) 196
D) 169
E) 120

Experts: only Veritas Prep experts, please!

Basically, you're picking two from a group of 16. And the order doesn't make a difference. Whether it's the Mets playing the Yankees or the Yankees playing the Mets -- it's the same game, right? So it's a combinations question. Basically do

16x15/2x1
=120

If you don't understand why it's done this way, let's look at how it would be done with a much smaller group, let's say only 5 teams.

ABCDE

If we just did every pairing it'd be
AA AB AC AD AE
BA BB BC BD BE
CA CB CC CD CE
DA DB DC DD DE
EA EB EC ED EE

That would be 5x5, but of course, that's crazy -- team A can't play itself.

So it's more like 5x4, not 5x5....

AB AC AD AE
BA BC BD BE
CA CB CD CE
DA DB DC DE
EA EB EC ED

But of course only half of those are distinct games. AB is the same as BA, and AC is the same as CA, etc.

AB AC AD AE
BA BC BD BE
CA CB CD CE
DA DB DC DE
EA EB EC ED

So it's like 5x4, but divided by two to get rid of the "doubles."
5x4/2 = 10.

BTW...
I'm not a Veritas Expert but that's a random qualification. I'm not sure why you need a Veritas expert to solve this relatively straightforward combinations question.

I very much agree with your explanation, except that the Yankees don't have anything to do with soccer;)

OA is E.

Correct answer: (E)

Solution: For any particular game, we have 16 options for the first team and 15 options for the second. 16 * 15 = 240. For our arrangements, though, Team A against Team B is the same as Team B against Team A, so we have to divide by two to get rid of the duplicates. 240/2=120, or answer (E).

Alternatively, because the problem is essentially asking how many groups of 2 teams can be generated from a pool of 16 teams, we can use the combination formula: N!/([K!][N-K]!). Our N (the number of things to be combined) is 16, and our K (the number of things in each group) is 2. Using these values generates the same answer, 120.

Anytime that you are working with selecting two from any number you can figure out just through a little hard work. Each team can play every other team. So team A has 15 opponents - then team B has only 14 opponents because A has played B already. So you can just take the total number of teams - 1 (in this case 15) and add each decreasing number so 15+14+13+12+11+10...+1 = 120. Not the most elegant way but sometimes on the test you are stuck and need to do something.

Well, if the OA is A and the source is Veritas Prep, then you might wanna change that:)

David@VeritasPrep wrote:OA is A

Correct answer: (E)

Solution: For any particular game, we have 16 options for the first team and 15 options for the second. 16 * 15 = 240. For our arrangements, though, Team A against Team B is the same as Team B against Team A, so we have to divide by two to get rid of the duplicates. 240/2=120, or answer (E).

Alternatively, because the problem is essentially asking how many groups of 2 teams can be generated from a pool of 16 teams, we can use the combination formula: N!/([K!][N-K]!). Our N (the number of things to be combined) is 16, and our K (the number of things in each group) is 2. Using these values generates the same answer, 120.

Anytime that you are working with selecting two from any number you can figure out just through a little hard work. Each team can play every other team. So team A has 15 opponents - then team B has only 14 opponents because A has played B already. So you can just take the total number of teams - 1 (in this case 15) and add each decreasing number so 15+14+13+12+11+10...+1 = 120. Not the most elegant way but sometimes on the test you are stuck and need to do something.

Great Work, Its always good to know different ways to reach the answer,
thanks.

Another method :
Suppose we have teams from A1 to A16
A1 will play games with 15 teams, A2 with 14.... and A15 with 1 (By this way we have taken no duplicates).
So, total games = 15 + 14 + 13 + .... + 2 + 1
And,
Sum of n numbers = ((n)*(n+1))/2
In this case = (15*16) / 2 = 120

For this type of problem, which is a combination problem but not a difficult one the quickest way is the conceptual approach

Since we have 16 teams, that means in one instance of all 16 teams playing there will be 8 games

Since 1 team will have to play all 15 other teams in the league, this 8 game instance will be played 15 times until all the teams have played eachother.

15 instances * (8 games/instance) = 120, pick E

16!/2!(16-2)!

= 16!/2!(14)!

=16*15/2

=240/2

=120

E??

David's expalantion is excellent as usual...

a similar method i picked up recently

using the SLOT method

a match can be defined as two available team partcipation in a match(defined in question too)
so we have two slots. First slot can be taken by any of the 16 teams. the next one by any of the remaining 15 teams. And in this case order does not matter(A vs B is the same as B vs A) hence we need to divide by 2!

16X15/2!=120

16 teams. Each team plays against each other team.
Thus no. of matches = no. of ways to select 2 teams (select 2 teams and conduct a match)
=16C2 = 120
IMO E

IMO E

16
C
2
That leads to 120 .

I was also thinking if each team engages with 8 other team -thats 15 * 8 =120 .

Yeah 15*16/2 =120
Answer E

yeah the answer to this question is option E.it is a combination problem as order of teams does not matter