Beginners Problem 002

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sakibtanvir
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Beginners Problem 002

Unread post by sakibtanvir » Sat Feb 04, 2012 10:36 am

In the 'Barkclays English Premier league' the total of the points of all teams is $420$ .There were $2$ points for a win,1 point for a draw and no point for lose.If each team played twice against each team,Then what was the number of the leagues participated in the competition?
I solved this in 5 minutes.You guys are smarter than me.Surely you will solve in a minute or something.
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nafistiham
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Re: Beginners Problem 002

Unread post by nafistiham » Sat Feb 04, 2012 3:28 pm

If one knows about combination.there is nothing to
In any match, there were $2$ points given, and each pair played twice.so, if we suspect the number $k$,
\[_{}^{k}\textrm{C}_2=\frac{420}{4} \]
\[\Rightarrow k=15\]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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sakibtanvir
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Location:24.4333°N 90.7833°E

Re: Beginners Problem 002

Unread post by sakibtanvir » Sat Feb 04, 2012 7:20 pm

My solution was like that:
Consider the number of teams as $n$.Now the number of matches will be \[\frac{n(n-1)}{2}*2\]
\[=n(n-1)\]
let the number of wins $w$ .
So,\[2w+n(n-1)-2w=420\]
\[\Rightarrow n(n-1)=420\]
Now solving the quadratic we get n=21.
The answer is 21. :lol:
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.

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nafistiham
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Re: Beginners Problem 002

Unread post by nafistiham » Sun Feb 05, 2012 2:44 pm

notice that in each match a total of $2$ points was achieved no matter what happened.so
\[n(n-1) \cdot 2 = 420\]
\[\rightarrow n(n-1)=210\]
\[\rightarrow n=15\]
;)
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

sakibtanvir
Posts:188
Joined:Mon Jan 09, 2012 6:52 pm
Location:24.4333°N 90.7833°E

Re: Beginners Problem 002

Unread post by sakibtanvir » Sun Feb 05, 2012 7:00 pm

But the draw should be considered also. :?: :?:
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.

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nafistiham
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Re: Beginners Problem 002

Unread post by nafistiham » Sun Feb 05, 2012 8:32 pm

suppose team a and team b are playing.there are $3$ cases. a wins, a loses or a draws. whatever happens the total point of a and b after this match will be $2$. that's what i am trying to say.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

sakibtanvir
Posts:188
Joined:Mon Jan 09, 2012 6:52 pm
Location:24.4333°N 90.7833°E

Re: Beginners Problem 002

Unread post by sakibtanvir » Mon Feb 06, 2012 12:30 pm

Ohhooooooo,sorryyyyy :oops: :oops: :oops: :oops:
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.

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