If there are 12 persons in a party and if each two of them shake

With 12 people , we need to choose a subset of two different people where order does not matter. Also, we need to choose all such subsets because each person is shaking hands with everyone else exactly once. The number of ways is: ${ }^{n} \bar{C}_{r}$

Where: $n=12 \& r=2$

${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{12} \mathrm{C}_{2}$

$\Rightarrow{ }^{12} C_{2}=\frac{12 !}{(12-2) ! \times 2 !}$

$\Rightarrow{ }^{15} C_{11}=\frac{12 !}{10 ! \times 2 !}$

$\Rightarrow{ }^{15} C_{11}=\frac{12 \times 11 \times 10 !}{10 ! \times 2 !}$

$\Rightarrow{ }^{15} \mathrm{C}_{11}=\frac{12 \times 11}{2 \times 1}$

$\Rightarrow{ }^{15} \mathrm{C}_{11}=66$

Ans: In total 66 handshakes are possible, if there are 12 persons in a party and if each two of them shake hands with each other