Question

In a statewide soccer competition, $36$ games were played.

Each team participating in the competition played once with every other team.

How many teams participated in the statewide competition?

Answer 1

Step-1: Find the numbers of teams participated in the statewide competition:

.Let the number of teams be $\mathrm{n}$.

Given that $36$ games were played.

Game can be played $2$ teams at a time.

Total games will be ${}^{\mathrm{n}}\mathrm{C}_2$.

Total number of games are $36$.

Hence, ${}^{\mathrm{n}}\mathrm{C}_2=36$.

Apply combination formula ${}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{r})!\times \mathrm{r}!}$.:

$$\begin{gathered} {}^{\mathrm{n}}\mathrm{C}_2=36 \\ \frac{\mathrm{n}!}{(\mathrm{n}-2)!\times 2!}=36 \\ \frac{\mathrm{n}!}{(\mathrm{n}-2)!\times 2!}=36 \end{gathered}$$.

Apply the identity $\mathrm{n}!=n(n-1)(n-2)!$:

$$\begin{gathered} \frac{\mathrm{n}.(\mathrm{n}-1)(\mathrm{n}-2!}{(\mathrm{n}-2)!\times 2!}=36 \\ \frac{\mathrm{n}(\mathrm{n}-1)}{2!}=36 \\ \frac{{\mathrm{n}}^2-\mathrm{n}}{2}=36 \\ {\mathrm{n}}^2-\mathrm{n}=36\times 2 \\ {\mathrm{n}}^2-\mathrm{n}=72 \\ {\mathrm{n}}^2-\mathrm{n}-72=0 \end{gathered}$$.

This is an algebraic equation .

Step-2: Simplify the equation.

$$\begin{gathered} {\mathrm{n}}^2-\mathrm{n}-72=0 \\ {\mathrm{n}}^2-9\mathrm{n}+8\mathrm{n}-72=0 \\ \mathrm{n}(\mathrm{n}-9)+8(\mathrm{n}-9)=0 \\ (\mathrm{n}-9)(\mathrm{n}+8)=0 \\ \mathrm{n}-9=0,\mathrm{n}+8=0 \\ \mathrm{n}=9,\mathrm{n}=-8 \end{gathered}$$

Ignore $-8$ because number of teams can not be negative.

Number of teams are $9$.

Hence, $9$ teams participated in the statewide competition.