Permutation-Combination
Aptitude

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Q.

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:

 A.

$18$

 B.

$19$

 C.

$17$

 D.

$16$

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Solution:
Option(A) is correct

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

$={^nC_2}$

$= 153$

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.


(5) Comment(s)


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.



Snehal
 ()

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



Abhishek
 ()

I need clear solution for this



Chandrasekaran
 ()

I need clear solution for this


RAKESH
 ()

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

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17=153.

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

17+1=18.