Question

On the occasion of Deepawali festival, each student of a class sends greeting cards to the other. If a total of $306$ greeting cards are exchanged, then the number of students in the class is?

Answer 1

We will first discuss the logic and plan with which we can count the number of greeting cards exchanged in the class with the help of the concepts of permutations and combinations. Then, we form a mathematical expression using the number of students in the class as a variable. Then, we expand the combination formula as $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ and find the required answer.

Complete step by step solution:
We are given that each student of a class sends greeting cards to others and are provided the total number of cards exchanged as $306$.
We know that the greeting card is exchanged between any two students in class.
So, we just have to find the number of pairs in the class.
Let us assume the total number of students in a class is $n$.
Then, we can find the number of pairs using the combination formula $^n{C_2}$.
But, we also know that two greeting cards are exchanged between one pair of students. So, we multiply the number of pairs in the class by two to get the number of greeting cards exchanged.
So, we get,
$\Rightarrow 2{ \times ^n}{C_2} = 306$
Dividing both sides by two, we get,
${ \Rightarrow ^n}{C_2} = \dfrac{{306}}{2}$
${ \Rightarrow ^n}{C_2} = 153$
The simplified form of the mathematical expression $^n{C_r}$ is $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ .
Here the term $n!$ defines the notion of multiplication of first n natural numbers.
This means $n! = 1 \times 2 \times 3 \times ....n$.
So, we have, $^n{C_2} = \dfrac{{n!}}{{2!\left( {n - 2} \right)!}}$.
Hence, we get,
$\Rightarrow \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} = 153$
Cancelling out the $\left( {n - 2} \right)!$ common in both numerator and denominator and putting the value of $2!$, we get,
$\Rightarrow \dfrac{{n \times \left( {n - 1} \right) \times \left( {n - 2} \right)!}}{{2\left( {n - 2} \right)!}} = 153$
$\Rightarrow \dfrac{{n \times \left( {n - 1} \right)}}{2} = 153$
$\Rightarrow {n^2} - n = 306$
$\Rightarrow {n^2} - n - 306 = 0$
Now, we solve the given quadratic equation using the splitting the middle term. So, we get,
We split the middle term $- n$ into two terms $- 18n$ and $17n$ since the product of these terms, $306{n^2}$ is equal to the product of the constant term and coefficient of ${n^2}$ and sum of these terms gives us the original middle term, $- n$.
$\Rightarrow {n^2} - 18n + 17n - 306 = 0$
Taking the common term outside the bracket, we get,
$\Rightarrow n\left( {n - 18} \right) + 17\left( {n - 18} \right) = 0$
$\Rightarrow \left( {n + 17} \right)\left( {n - 18} \right) = 0$
Now, either $\left( {n + 17} \right) = 0$ or $\left( {n - 18} \right) = 0$
So, we get, $n = - 17$ or $n = 18$
Since the number of students in a class cannot be negative. So, we get, $n = 18$. Therefore, there are $18$ students in the class.

Note:
There are some constraints in the form of $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ such as the value of r cannot be negative and greater than n. Also, we need to remember the fact that the notion of choosing r objects out of n objects is exactly equal to the notion of choosing $(n-r)$ objects out of n objects. The mathematical expression is $^n{C_{\left( {n - r} \right)}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}{ = ^n}{C_r}$.