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In a hockey championship, there are $153$ matches played.

Every two team played one match with each other. The number of teams participating in the championship is:









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Option(A) is correct

Let there were $x$ teams participating in the games, then total number of matches:


$= 153$

On solving, we get $n= -17$ and $n=18$.

It cannot be negative so $n = \textbf{18}$ is the answer.

Edit: For an alternate explanation, check comment by RAKESH.

(6) Comment(s)

Brandon H

A more algebraic approach...

Total # of combinations is xC2 or "x Choose 2" which must equal 153. --->

x! / [(x-2)! * 2!)] = 153

Multiply both sides by 2! --->

x! / (x-2)! = 306

Keep in mind, that x! is x * (x-1) * (x-2) * (x-3) * (x-y)....etc. until (x-y) = 1

So in the case of x! / (x-2)! , all of the numbers after (x-1) will cancel out (they are being divided) and leave us with --->

x (x-1) = 306

From here just use algebra to create a quadratic polynomial

x^2 - x - 306 = 0

From here, factor and you'll get (find 2 numbers with a product of 306 and a sum of -1) --->

(x-18)(x+17) = 0

Solving for the zeros give us +18 and -17

-17 is obviously an answer that doesn't make logical sense in this case so the answer must be 18 teams.

Raj Singh

Rakesh's approach is to add the total no. of matches played, since it's given that total no. of matches played is 153. First team plays 17 matches, Second team plays with 16 matches without repetition, third team plays 15 matches without repetition... and so on.


Rakesh your method is great, however with regards to the reasoning, could you please explain again logic behind the adding of natural numbers?


I need clear solution for this


I need clear solution for this


here, compulsory to play two team one match so that, we will add the natural number till we get value 153. so our result will be like a


so by this we have to add value upto 17 and the one more team should add that is oppent team so the total number of team is