In the Mundane Goblet competition, 6 teams compete in a

Question

In the Mundane Goblet competition, 6 teams compete in a “round robin” format: that is, each team plays every other team exactly once. A team gets 3 points for a win, 1 point for a tie (a draw), and 0 points for a loss. What is the difference between the maximum total points and the minimum total points that can be gained by all teams (added together) in the Mundane Goblet competition?

(A) 15

(B) 30

(C) 45

(D) 60

(E) 75

(A) 15
(B) 30
(C) 45
(D) 60
(E) 75

Bunuel · Accepted Answer

There will be  C^2_6=15  games needed so that each team to play every other team exactly once ( C^2_6=15  is the # of ways we can pick two different teams to play each other).

Now, in one game max points (3 points) will be obtained if one team wins and another looses and min points (1+1=2 points) will be obtained if there will be a tie. Hence, maximum points that can be gained by all teams will be  15 games * 3 points=45  and the minimum points that can be gained by all teams will be  15 games * 2 points=30 , difference is  45-30=15 .

Answer: A.

Hope it helps.

KarishmaB · Answer

To add to what Bunuel said above, we are interested in the total points obtained by all the teams together i.e. the total points earned in all the matches. In any one match, if one team loses and gets 0, the other team wins and gets 3. So, in that match, a total of 3 points are generated. On the other hand, if the two teams draw, they both get a point each and in that game, a total of 2 points are generated. No game can generate less than 2 points. So the minimum points generated by a match are 2 and the maximum are 3. Hence we do 3*15 - 2*15.

zisis · Answer

No OA yet: my method:

max points: 
A B C D E F
15 12 9 6 3 0 = 45POINTS

min points 
A B C D E F
5 5 5 5 5 5 = 30

45-30= 15 points ====> (A)

vannbj · Answer

Maybe this is a dumb question but I have to ask. Is  C^2_6  the same thing as 6!/(2!*4!)? That’s the only way I could get the 15 by which you multiplied 3 and 2 and then subtracted 30 from 45.

kidchaos · Answer

Its just 3(total teams - 1) - 0
Coz max pts are if 1 team won all the matches & min pts are 0

In This case 3(6-1) - 0 = 15 - 0 = 15

HTH

galiya · Answer

"In the Mundane Goblet competition, 6 teams compete in a "round robin" format: that is, each team plays every other team exactly once. A team gets 3 points for a win, 1 point for a tie (a draw), and 0 points for a loss. What is the difference between the maximum total points and the minimum total points that can be gained by all teams (added together) in the Mundane Goblet competition?

*15
*30
*45
*60
*75
the decision:
6!/(2!*4!)=15 games
Maximum=3*15=45
Minimum=2*15=30, so
45-30=15
the correct answer is A"

Could somebody explain me please why the minimum score i had to calculate as 2*15, instead of 0*15???
I solved it as 3*15-0*15=45...or i just missed that there are no any earned scores when the team loses the game? :roll:

nammers · Answer

When a team wins one team gets 3 points and the other 0 meaning 3 points are earned between the two teams in the match.
When the teams draw both teams get a point each meaning 2 points are earned between the teams in the match (hence 2*15).

blink005 · Answer

good point nammers
the points are split between the two teams, and the question says together
so it's 2 i.e. a tie

Spidy001 · Answer

the question talks about total points of all teams - meaning they are not just interested in any specific team.

every match is played between two teams -
 two possible scenarios - one team wins and other loses => total points in that match = 3+0 = 3
 its tie - each team gets point => total points in that match = 1+1 = 2

so maximum possible points in any match = 3 and minimum =2

altogether we have 15 games, all 15 can fall in one of those scenarios above.

That's why maximum possible points = 15*3 = 45
 minimum possible points = 15*2 = 30

difference is 15.

sanjoo · Answer

Why we r not taking minumun value 0??

Bunuell..what actually i was doing..i multiplied 3*15=45..

and 15*0 =0...45-15=30..

what i think because if we take 0?? then one team loss is another team win..so it will take same 45 points ...

my concept behind it is rite??

Thanks

Bunuel · Answer

How can we have zero points in one game? The minimum points in one game is 1+1=2 points, when there is a tie between the teams.

sanjoo · Answer

Thanks alot Karishma and Bunuel

U both rocked

..

IanElliott · Answer

There will be  C^2_6=15  games needed so that each team to play every other team exactly once ( C^2_6=15  is the # of ways we can pick two different teams to play each other).

Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:

Bunuel I am not familiar at all with the notation you used (  C^2_6=15  )

Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.

In my case, I used n(n-1)/2 which seemed to work okay. Is one method preferable to the other?

Hope my question was clear.

danidan · Answer

C^2_6=15 is a quick notation to say: a combination (order doesn't matter) of 2 items in a pool of 6 items In order words: how many possibilities to pick 2 different teams among 6 teams More generally C^r_n is a combination of r items in a pool of n items Which leads to the general formula: n!/(n-r)!r! :arrow:

6!/(2!4!)

PrashantPonde · Answer

IanElliott, check below link about Combinatorics topic and practice problems. You may find it useful to understand above concepts. math-combinatorics-87345.html

KarishmaB · Answer

There are various ways to get a particular answer. People choose a particular method depending on how they think.

Your thought would have been something like this: There are 6 teams. The first team plays with every other team i.e. with 5 teams and goes away.
Now, the second plays with the rest of the 4 teams and goes away. Now the third team plays with the rest of the 3 teams and so on...

So total there are 5 + 4+ 3 + 2 + 1 and you used n(n-1)/2 here.

Another way to think about it is this: There are total 6 teams. Every pair of two teams played one match. Out of 6 teams, we will try to make as many distinct pairs of 2 teams each as possible. For this, we use 6C2 = 6!/2!*4!

There is absolutely nothing wrong with either one of the methods. But you must be able to think in nCr terms too because it comes in very handy in many questions. nCr = choose r items from n

pavanpaone · Answer

Hi ,the method is called permutations and combinations.here we want to find the total number of games played by all teams together.Here we have to choose 2 teams from 6 teams to play game i.e here we are using combinations so the formula to find out the total combination how each team can play with other exactly ones the formula is (n C r) n-represents total teams ,C - combination and r - represents minimum no.of teams required to play a game .In this problem n=6 and r= 2

n C r = n! /( (n-r)! * r !) = (1*2*...*n)/((1*2.....(n-r))*(1*2.....r)) .here 6!/(4! * 2!) = 15.

I hope you understood

pavanpaone · Answer

total no of games played = 15
max points gained by all teams together = 15* 3 = 45
minimum no.of points gained by all teams together is = 15*2 =30
 difference between max and min points = 45-30=15

The procedure to calculate min number of points:

here we require the minimum number of points can be gained by all teams together is 2 points i.e if the game is tied each time will get 1 points making min points each game is 2 points so if all the games are tie ,the minimum points of all games is = 15* 2=30

cumulonimbus · Answer

Hi Karishma,

Can you please explain why did we add = 5+4+3+2+1 here?

In the Mundane Goblet competition, 6 teams compete in a

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