Question 1087253
<font color="black" face="times" size="3">There are 16 students. Any given student has 15 people to shake hands with. Doing this 16 times means we have 16*15 = 240 orderings. 


Keep in mind that instructions state <font color=blue>"Each shakes hands with each other exactly once"</font>. 


So we need to get rid of the duplicate handshakes. For example, let's say we have student A and student B. One handshake is AB while another handshake is BA. The order doesn't matter as it's the same group. 


To correct this, we divide by 2 and it gets us 240/2 = 120 which is the final answer.


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Here is a table showing all the possible handshakes. Along the left and top borders are the student names (code name letters A through P, which is the 16th letter of the alphabet). 


<img src="https://i.imgur.com/JVUD2ex.png">


Combos like BB or CC are impossible so we ignore anything along the main diagonal. Those are marked with X. Anything below the diagonal is a mirror of what is above the diagonal. This is where the double-counting comes in. So we simply pick one side. I'm going to pick the stuff above the diagonal. Anything below the diagonal will have X in it. The boxes with S in it indicate a handshake. If you carefully count out the S letters, you'll find that there are 120 of them. An alternative is to sum the row counts to get: 15+14+13+...+3+2+1 and you'll get 15*(15+1)/2 = 15*16/2 = 120 which is the same as before. 
Using the formula is the preferred method since the table is a bit tedious. </font>