Question

1. there will be a dance party where people only dance in pairs and everyone will dance with everyone else once. find the minimum number of people required to be at the party if at least 3,240 dances to occur.

Explanation:

Step1: Define the combination formula

The number of pairs (dances) formed from \( n \) people, where each pair dances once, is given by the combination formula \( C(n, 2)=\frac{n(n - 1)}{2} \), since we are choosing 2 people out of \( n \) for each dance and the order of the pair does not matter.

Step2: Set up the inequality

We want the number of dances to be at least 3240, so we set up the inequality:

$$ \frac{n(n - 1)}{2}\geq3240 $$

Multiply both sides by 2 to get:

$$ n(n - 1)\geq6480 $$

Which expands to:

$$ n^{2}-n - 6480\geq0 $$

Step3: Solve the quadratic equation

We first solve the quadratic equation \( n^{2}-n - 6480 = 0 \) using the quadratic formula \( n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 1 \), \( b=- 1 \) and \( c=-6480 \).

$$ n=\frac{1\pm\sqrt{(-1)^{2}-4\times1\times(-6480)}}{2\times1}=\frac{1\pm\sqrt{1 + 25920}}{2}=\frac{1\pm\sqrt{25921}}{2} $$

We calculate \( \sqrt{25921}\approx161 \) (since \( 161^{2}=25921 \)). So the roots are:

$$ n=\frac{1 + 161}{2}=81,\quad n=\frac{1-161}{2}=-80 $$

Since the number of people \( n \) must be positive, we consider the positive root. The quadratic function \( y=n^{2}-n - 6480 \) opens upwards (because \( a = 1>0 \)), so the inequality \( n^{2}-n - 6480\geq0 \) is satisfied when \( n\geq81 \) or \( n\leq - 80 \). Since \( n \) represents the number of people, we take \( n\geq81 \). We can check for \( n = 80 \): \( \frac{80\times79}{2}=\frac{6320}{2}=3160<3240 \). For \( n = 81 \): \( \frac{81\times80}{2}=\frac{6480}{2}=3240 \), which satisfies the condition of at least 3240 dances.

Answer:

81