How many Handshakes?

This is a question of the number of ways to choose $2$ persons out of $40$ . It answer is looking for the number of combinations, $N={ _{ 40 }{ C }_{ 2 } }\space or\space \left( \begin{matrix} 40 \\ 2 \end{matrix} \right) =\frac { 40! }{ 2!\times 38! } =\frac { 40\times 39 }{ 2\times 1 } =\boxed{780}$ .

Method 1 - The number of ways in which people can shake hands is 40C2 i.e.780 which includes the handshake done by A with B and excludes handshake done by B with A. Method 2 - 1st person:39 (including just handshakes done by him/her not offered to him/her) 2nd person:38 (including just handshakes done by him/her not offered to him/her) ........ 39th person:1 (including just handshakes done by him/her not offered to him/her) 40th person:0 (including just handshakes done by him/her not offered to him/her) Add them up to get the total no. of handshakes : 39+38+...+1 i.e. 39*40/2 =780

No. of shake hands = n(n – 1)/2 where n total persons

40 × 39/2 = 780

First person will shake hands with 39 persons,then second person can shake with 38 persons like this way third person 37...so on. Now add all of them>>>39+38+37.....5+4+3+2+1=780.