HANDSHAKES

If N people at the party, then if we line the people up in a row, the first person shakes hands with the N-1 people to his left, the second person shakes hands with the N-2 people to his left, until the second to last person in the row shakes hands only with the last person in the row. Hence the number of handshakes is the sum of the integers from 1 to N-1. It can be shown that this sum is (N-1) N/2. Hence (N-1) N/2=105 or equivalently (N-1)*N=210. Since 14^2<210<15^2, we see that N must be 15 and hence the total number of hangs is 30.

Consider there were $x$ people in the party. So if every person shakes hand with every person, then the total handshakes will be ${ x }^{ 2 }$ or $x\times x$

But since people don't shake hand with themselves, every person shakes hand with $x - 1$ people. Therefore total handshakes = $x\times (x-1)$

But now it's given that no one did handshake twice with one person . But with the above formula, every person shakes hand with every other twice.

Therefore, on only shaking hands with one person, handshakes = $\frac { x\times (x-1) }{ 2 }$

According to the question, $\frac { x\times (x-1) }{ 2 } =105$

So now we can solve the equation -

$\frac { x\times (x-1) }{ 2 } =105\\ x\times (x-1)=210\\ { x }^{ 2 }-x=210\\ { x }^{ 2 }-x-210=0$

On solving the quadratic equation, we find that $(x+14)(x-15)=0$

We can omit the negative value. So $x = 15$ or number of people at party = 15. But since in the question it's asked to find the number of hands, it's $15\times2 = \boxed{30}$

If n is the number of man then total number of handshakes is =nC2=105 Or n^2-n-210=0 Solving this equation we have
n=15 and -14 Here n= -14 is not acceptable so the total number of hands is= 30