Question

There are $12$ persons at a party, and if each two of them shake hands with each other, how many handshakes happen at the party?

Answer 1

A number of handshakes happen at the party.

The combination is the number of ways of selecting objects such that the order of selection is not to be counted.

Combination Formula:

${}_n{C}_r=\left(\begin{array}{l}n\\ r\end{array}\right)=\frac{{}_n{P}_r}{r!}=\frac{n!}{r!(n-r)!}$

The total number of handshakes is the same as the number of ways of selecting $2$ persons among $12$ persons i.e.,

$$\begin{gathered} \Rightarrow \mathrm{Number}\mathrm{of}\mathrm{hanshake}{=}^{12}{\mathrm{C}}_2 \\ =\frac{12\times 11}{2}[\because {}^n{C}_r=\frac{n!}{r!(n-r)!}=\frac{n(n-1)(n-2)\dots \dots (n-r+1)}{r!}] \\ =66 \end{gathered}$$

Hence, $66$ handshakes happen at the party.