Handshakes on a plane - Corrected version

To solve this problem, you need to know a couple things. First off is that a handshake needs two hands. You have 200 hands to choose to start. Then there are 199 hands to choose for the second. 200 * 199 = 39800. But you're not done. You divide by 2! because you are overcounting. 39800 divided by 2 is 19900.

Passenger One or P1 shakes hands with P2 ~ P200, a total of 200 – 1 = 199 other passengers. But all of them shake hands with each other, not just P1 alone who did so, therefore 200 passengers shook 199 other hands for a total of 200 × 199. But remember that we counted the handshakes twice from both participants' POVs (eg. P1 🤝 P200 from P1's perspective and P200 🤝 P1 from P200's perspective, but it really was the same event, NOT a home and away game here). So answer is 200 × 199 / 2 = 19900.

It takes two persons to have a handshake. Therefore, the total number of handshakes among $n$ persons, $N(n) = \dbinom n2 = \dfrac {n(n-1)}2$ . For $n=200 \implies N(200) = \dfrac {200 \cdot 199}2 = \boxed{19900}$ .

Summation formula: (200-1)+(199-1)... by each passenger until goes to 1. (199*200)/2