Question

In the football championship, 153 matches were played. Every team played one match with each other. The number of teams participating in the championship is-
A.17
B.18
C.9
D.None of these

Answer 1

Here, we will use the concept of Arithmetic Progressions to solve this question. We will use the given information to form a series representing the match played by the teams in the championship. Then we will use the sum formula to find the total number of teams that participated in the championship.

Formula Used:
In an Arithmetic Progression, ${S_N} = \dfrac{N}{2}\left( {a + {a_n}} \right)$, where $N$ is the total number of terms, $a$ is the first term and ${a_n}$ is the last term.

Complete step-by-step answer:
Let the total number of teams be $N$ namely, ${T_1},{T_2},{T_3},....{T_N}$.
Now, according to the question, every team played one match with each other.
Hence, by using permutations, we know that there are a total of $N$ teams, hence, ${T_1}$ can play matches with $\left( {N - 1} \right)$ teams.
This is because, ${T_1}$ cannot play with itself, hence, we will exclude this case.
Now, ${T_2}$ can play $\left( {N - 2} \right)$ games, keeping in mind the fact that we cannot include ${T_2}$ itself and one match has already been played and included in ${T_1}$.
Similarly, we will go on by subtracting 1 game each time from the total possible games.
Hence, we will get the Arithmetic Progression as:
$\left( {N - 1} \right),\left( {N - 2} \right),\left( {N - 3} \right),....0$
Hence, total number of possible games being played will be: $\dfrac{{N\left( {N - 1} \right)}}{2}$
Hence, substituting the first term, $a = \left( {N - 1} \right)$ and the last term ${a_n} = 0$ in the formula ${S_N} = \dfrac{N}{2}\left( {a + {a_n}} \right)$, we get,
${S_N} = \dfrac{N}{2}\left( {N - 1 + 0} \right)$
$\Rightarrow {S_N} = \dfrac{{N\left( {N - 1} \right)}}{2}$
But, according to the question, in the football championship, 153 matches were played
Hence, we get,
$\Rightarrow \dfrac{{N\left( {N - 1} \right)}}{2} = 153$
Multiplying the terms, we get
$\Rightarrow {N^2} - N = 153 \times 2 = 306$
$\Rightarrow {N^2} - N - 306 = 0$
Now, the above equation is a quadratic equation so we will factorize it to find the required value.
Splitting the middle terms, we get
$\Rightarrow {N^2} - 18N + 17N - 306 = 0$
$\Rightarrow N\left( {N - 18} \right) + 17\left( {N - 18} \right) = 0$
Factoring out common terms, we get
$\Rightarrow \left( {N + 17} \right)\left( {N - 18} \right) = 0$
Using zero product property, we get
$\begin{array}{l}\left( {N - 18} \right) = 0\\\ \Rightarrow N = 18\end{array}$
Or
$\begin{array}{l}\left( {N + 17} \right) = 0\\\ \Rightarrow N = - 17\end{array}$
But, the number of teams can never be negative, so we will reject the negative value.
Therefore, the total number of teams participating in the championship is 18.
Hence, option B is the correct answer.

Note: Here, we have used Arithmetic Progressions to form the equation. An Arithmetic Progression is a sequence of numbers such that the difference between any term and its preceding term is constant. This difference is known as the common difference of an Arithmetic Progression (AP). A real life example of AP is when we add a fixed amount in our money bank every week. Similarly, when we ride a taxi, we pay an amount for the initial kilometer and pay a fixed amount for all the further kilometers, this also turns out to be an AP.