Question

In a huge tournament, $45$ matches were played. Find out how many people are involved if it is known that each participant played one game with every other participant in the tournament.

Answer 1

Step 1: Find the number of participants.

It is given that total matches are $45$ and each participant played one game with every other participant in the tournament .

Let total participants be $n$.

Now, we can say that the first participant played with $(n-1)$ participants.

Similarly second participant participated with $(n-2)$ participants.

Third participant participated with $(n-3)$ participants

and so on…

Continuing this we got an arithmetic progression as,

$n-1,n-2,n-3,...,3,2,1$

Step 2: Sum of arithmetic progression.

We know that, sum of natural numbers is $n+\left(n-1\right)+\left(n-2\right)+...+3+2+1=\frac{n\left(n+1\right)}{2}$

So, we have

$\begin{array}{rcl}\left(n-1\right)+\left(n-2\right)+...+3+2+1& =& \frac{n\left(n+1\right)}{2}-n\\ & \Rightarrow & \left(n-1\right)+\left(n-2\right)+...+3+2+1=\frac{n\left(n+1\right)-2n}{2}\\ & \Rightarrow & \left(n-1\right)+\left(n-2\right)+...+3+2+1=\frac{n^2+n-2n}{2}\\ \therefore \left(n-1\right)+\left(n-2\right)+...+3+2+1& =& \frac{n^2-n}{2}\end{array}$

According to the question $\left(n-1\right)+\left(n-2\right)+...+3+2+1=45$.

Then we have, $\begin{array}{l}\frac{n^2-n}{2}\end{array}=45$.

Step 3: Solve for $n$.

$\begin{array}{rcl}\begin{array}{l}\frac{n^2-n}{2}\end{array}& =& 45\\ & \Rightarrow & n^2-n=45·2\left[\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}2\right]\\ & \Rightarrow & n^2-n=90\\ & \Rightarrow & n^2-n-90=0\left[\mathrm{Subtracting}\mathrm{both}\mathrm{sides}\mathrm{by}90\right]\\ & \Rightarrow & n^2-10n+9n-90=0\\ & \Rightarrow & n\left(n-10\right)+9\left(n-10\right)=0\\ & \Rightarrow & \left(n-10\right)\left(n-9\right)=0\\ \therefore n& =& 10,9\end{array}$

But if $n=9$,

$9+8+7+6+5+4+3+2+1=35\ne 45$

$\therefore n=10$

Hence, the number of people involved are $10$.