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## Homework Statement

The question is to find the number of people at a party, if every two people shake hands and the number of handshakes is 1275.

## Homework Equations

## The Attempt at a Solution

Here is my partial solution:

After having performed simple cases of the problem--when there are 2, 3, 4, and 5 people--, I found that if there are n people, then first person chooses to shake hands with n - 1 people.

The number of handshakes can be written as

(n-1) + (n-2) + (n-3) +...+ 1 + 0 = 1275 (the zero connotes the fact that the last person does not choose to shake hands with anyone.) If we could somehow collect the n-terms so that it takes the form of an, where a is a some function (either a constant or some function of n)

Let's now investigate the relationship between the number of people and the number of handshakes.

# of people......................# of handshakes

.....2..................................1 = 1

.....3.............................2 + 1 = 3

.....4........................3 + 2 + 1 = 6

.....5................. 4 + 3 + 2 + 1 = 10

Looking at the n = 5 equation, 4 the is actually an n term; it's n - 1, with n = 5; and so isn't 3,

n - 2, as well as 2 and 1, n - 3 and n - 4, respectively. So, there are 4 terms with n them, or there are n - 1 terms. So, the number of handshakes is n(n-1). But this is an over count. This says that every person shakes hands with n-1 people. We know that each person shakes hands with one less person than the last. If we use the formula n(n-1) to calculate the number of handshakes, we get 5(5-4) = 20, which is double the actual number of handshakes. n(n-1) must be twice the number of handshakes. Hence, n(n-1)/2.

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As I look back at this, I feel like my arguments are a little vague. Is there a more mathematical argument to derive the formula n(n-1)/2, especially the 1/2 factor. I understand that this isn't a mathematically rigorous class, the intention is to stimulate thinking, but I would to see if there is in fact a way to derive the formula, and to see if I can understand.

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Here is another solution that I was unsure how to use:

Looking back at our table, we see the number of hand shakes is 1 = y2, 3 = y3, 6 = y4, 10 = y5.

y4 - y3 = 3 and y5 - y4 = 4

(y5 - y4) - (y4 - y3) = 4 - 3

y5 - y3 = 1

So, my conjecture is that [itex]y_n - y_{n-2} = 1[/itex] But I can't quite figure out how to use this fact.