Respuesta :
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,
[tex]=\ ^nC_2[/tex]
As they played 2 games with every other participant in the tournament.
So the total number of games is,
[tex]=\ 2\times ^nC_2[/tex]
But it is given to be 20, so
[tex]\Rightarrow \ 2\times ^nC_2=20[/tex]
[tex]\Rightarrow \ ^nC_2=10[/tex]
[tex]\Rightarrow \dfrac{n!}{2!(n-2)!}=10[/tex]
[tex]\Rightarrow \dfrac{n(n-1)}{2}=10[/tex]
[tex]\Rightarrow {n(n-1)=20[/tex]
As [tex]5\times 4=20[/tex], so we get n=5.
Therefore, 5 players participated in the tournament.
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 :
[tex]^nC_2[/tex]
We are also given that each participant played 2 games with every other participant.
So, total no. of games played =[tex]2 * ^nC_2[/tex]
Since we are given that total no. games played = 20
⇒[tex]2 * ^nC_2 = 20[/tex]
⇒[tex]^nC_2 = \frac{20}{2}[/tex]
⇒[tex]^nC_2 =10[/tex] --(a)
Formula of combination:
⇒[tex]\frac{n!}{r! * (n-r)!}[/tex]
So, solving (a) further using formula
⇒[tex]\frac{n!}{2! * (n-2)!}=10[/tex]
⇒[tex]\frac{n*(n-1)*(n-2)!}{2! * (n-2)!}=10[/tex]
⇒[tex]\frac{n*(n-1)}{2*1}=10[/tex]
⇒[tex]\frac{n^{2} -n}{2*1}=10[/tex]
⇒[tex]n^{2} -n=10*2[/tex]
⇒[tex]n^{2} -n=20[/tex]
⇒[tex]n^{2} -n-20=0[/tex]
⇒[tex]n^{2} -5n+4n-20=0[/tex]
⇒[tex]n(n-5)+4(n-5)=0[/tex]
⇒[tex](n-5)(n+4)=0[/tex]
⇒[tex](n-5) =0 , (n+4)=0[/tex]
⇒ n = 5 , n =-4
Neglect the negative value since no. of players cannot be negative.
Thus no. of player involved is 5.
