Question

$\overbrace{ \underbrace \orange{ \tt{ \bold{[Maths] }}}}$

At an annual function of a school, each student gives gift to every other student. If the number of gifts is 1980, find the number of students.​

Answer 1

Let the number of students be $x$

If each student gives gift to every other student, then it is the same as permuting those $x$ students into groups of 2.

So, we have,

$^xP_2 = 1980$

$\dfrac{x!}{(x - 2)!} = 1980$

$\dfrac{x(x - 1)(x - 2)!}{(x - 2)!} = 1980$

$x(x - 1) = 1980$

${x}^{2} - x - 1980 = 0$

${x}^{2} + 44x - 45x - 1980 = 0$

$x(x + 44) - 45(x + 44) = 0$

$(x + 44)(x - 45) = 0$

Hence, $x= -44, 45$

But, as the number of students can't be negative, Therefore $x= \boxed {45}$ are the total number of students.

Answer 2

Answer:

The above question is based on quadratic equation:-

Logic:-

As each student gives gift to every student but the last student did not give the gift as he is the last, so number of gifts are number of students-1.

Assumption:-

Let the number of students be x

Let the number of gifts be x-1

According to question:-

x(x-1)=1980

${x}^{2} - x = 1980$

${x}^{2} - x - 1980 = 0$

Product= -1980

Sum= -1

Numbers= -45,44

Splitting the mid term:-

${x}^{2} - 45x + 44x - 1980 = 0$

=x(x-45)+44(x-45)=0

=(x+44)(x-45)

x+44=0 or x-45=0

x= -44,45(number of students cannot be negative)

Therefore,Number of students=45 students

Number of gifts=45-1

=44 gifts