Question

If at the end of certain meeting, everyone had shaken hands with everyone else, it was found that 45 handshakes were exchanged, then the number of members present at the meeting, are

• A. 10
• B. 15
• C. 20
• D. 21

Hint:

Logic Tip: For simple permutation/combination problems like $n(n-1) = 90$, instead of writing out a full quadratic equation, simply ask yourself: "What two consecutive positive integers multiply to 90?" It's immediately obvious that $10 \times 9 = 90$, so the larger integer $n$ must be 10.

Answer 1

The Correct Option is A

Concept:
A handshake occurs between exactly 2 people. If there are $n$ people in a meeting and everyone shakes hands with everyone else exactly once, the total number of handshakes is equivalent to the number of ways to choose 2 people out of $n$. This is a basic combination problem given by the formula ${}^nC_2$.
Step 1: Set up the combinations equation.
Let $n$ be the total number of members present at the meeting. The total number of handshakes is given by: $${}^nC_2 = 45$$ Expand the combination formula $\frac{n!}{2!(n-2)!}$: $$\frac{n(n-1)}{2} = 45$$
Step 2: Solve the resulting quadratic equation.
Multiply both sides by 2 to clear the fraction: $$n(n-1) = 90$$ Expand the left side: $$n^2 - n = 90$$ $$n^2 - n - 90 = 0$$ Factor the quadratic equation by looking for two numbers that multiply to -90 and add to -1. Those numbers are -10 and 9: $$(n - 10)(n + 9) = 0$$
Step 3: Determine the valid value for n.
This gives two possible mathematical solutions: $$n = 10 \quad \text{or} \quad n = -9$$ Since $n$ represents the number of people, it cannot be a negative number. Therefore, we discard $n = -9$. The number of members present is 10.