Question

In a small chess tournament, 20 matches were played. Find out how many people were involved if it is known that each participant played 2 games with every other participant in the tournament.

Answer 1

Answer:

5 players participated in the tournament.

Step-by-step explanation:

In a small chess tournament, 20 matches were played.

Let us assume that n number of players participated in the tournament

As in each game 2 players play, so the number of ways they can play is,

$=\ ^nC_2$

As they played 2 games with every other participant in the tournament.

So the total number of games is,

$=\ 2\times ^nC_2$

But it is given to be 20, so

$\Rightarrow \ 2\times ^nC_2=20$

$\Rightarrow \ ^nC_2=10$

$\Rightarrow \dfrac{n!}{2!(n-2)!}=10$

$\Rightarrow \dfrac{n(n-1)}{2}=10$

$\Rightarrow {n(n-1)=20$

As $5\times 4=20$, so we get n=5.

Therefore, 5 players participated in the tournament.

Answer 2

Answer:

Step-by-step explanation:

Given :  

20 matches were played in a small chess tournament.

Each participant played 2 games with every other participant in the tournament.

To Find : how many people were involved?

Solution :

Let no. of players involved be n

Since we know that for every match there should be two players out of n

So, number of ways they can play :

$^nC_2$

We  are also given that  each participant played 2 games with every other participant.

So, total no. of games played  =$2 * ^nC_2$

Since we are given that total no. games played = 20

⇒$2 * ^nC_2 = 20$

⇒$^nC_2 = \frac{20}{2}$

⇒$^nC_2 =10$  --(a)

Formula of combination:

⇒$\frac{n!}{r! * (n-r)!}$

So, solving (a) further using formula

⇒$\frac{n!}{2! * (n-2)!}=10$

⇒$\frac{n*(n-1)*(n-2)!}{2! * (n-2)!}=10$

⇒$\frac{n*(n-1)}{2*1}=10$

⇒$\frac{n^{2} -n}{2*1}=10$

⇒$n^{2} -n=10*2$

⇒$n^{2} -n=20$

⇒$n^{2} -n-20=0$

⇒$n^{2} -5n+4n-20=0$

⇒$n(n-5)+4(n-5)=0$

⇒$(n-5)(n+4)=0$

⇒$(n-5) =0 , (n+4)=0$

⇒ n = 5 , n =-4

Neglect the negative value since no. of players cannot be negative.

Thus no. of player involved is 5.