The first person can be selected in one of seven ways. Having selected him, the second can be selected from the remaining six in six ways. So there would appear to be 7*6 ways of selecting the couples shaking hands.
But, x shaking y's hand is the same handshake as y shaking x's hand. Thus each handshake is conted twice, So the total number of handshakes is 7*6/2 = 21
10 times
For everyone to shake hands with everyone else, there are 21 handshakes. (below copied from my answer of a similar question) This is an arithmetic progression and can be solved with the equation Sn=(1+(n-1))(n-1)/2 where Sn is the total sum of handshakes for n people. NB: I have used n-1 instead of n in the equation for the sum of an arithmetic progression, because you're not really going to shake hands with yourself, so you don't include the nth term, in this case 7. Alternatively, you can solve this geometrically by drawing a heptagon, drawing lines between all the vertices, and then adding all the lines up.
72
First person shakes hands 19 times, second person 18 etc, a total of 190.
The two hands make an angle of 155 degrees (and 205 degrees).
4950 handshakes, that is the definite answer
12
4950
9 handshakes correct answer is 10
200? D:
45 handshakes
10
Everyone shakes hands with 4 other people. Since there are 5 people in the room this would suggest there are 5*4 = 20 handshakes. However, you would then be double counting handshakes: A shaking hands with B and B shaking hands with A is, in reality, only one handshake. Thus there are 5*4/2 = 10 handshakes in all.
You could never guarantee 1000 handshakes because the people may choose not to shake hands! If each person did shake hands with everyone else, then 46 people would suffice.
If there are seven people, then the number of handshakes is 7*6/2 = 21
So, there will be 3 handshakes among the 3 people at the party.
6