Question 452441

So let's ay the room has only A and B---two people.

there is one handshake
(A with B).

Three people (A, B, and C), there are three handshakes
(A with B and C; B with C).

Now, let's try four people (A, B, C, and D), there are six handshakes
(A with B, C, and D; B with C and D; C with D).

So if we have n-1 people, the number of handshakes is the sum of the first n numbers. This is n(n-1)/2


a) If there are 5 people in the room, how many handshakes have taken place?


{{{n(n-1)/2=5(5-1)/2=5*4/2=5*2=10}}}


b) If there are 6 people in the room, how many handshakes have taken place?

{{{n(n-1)/2=6(6-1)/2=6*5/2=3*5=15}}}


c) If there are 7 people in the room, how many handshakes have taken place?

{{{n(n-1)/2=7(7-1)/2=7*6/2=7*3=21}}}

d) If there are n people in the room, how many handshakes have taken place? 

Write a recursive formula and explicit formula.

{{{n(n-1)/2}}}

e) Where are these numbers on Pascal's Triangle?

these are triangular numbers

f) If there were 190 handshakes altogether, how many people were in the room?

{{{n(n-1)/2=190(190-1)/2=190*189/2=17955}}}


Another way to think of this, if you have done some combinations, is for {{{6}}} people, in how many ways can you pick two. This is {{{6C2 = 6*5/2= 15}}}

for {{{190}}} people{{{190C2 = 190*189/2= 17955}}}