Word Problem

kmring

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If there are 100 people and each person shakes hands only one time with each person how many handshakes would that be?

The answer is 4950. My question is how do I put this into an algebra formula?

# People # Handshakes

1 0
2 1
3 3
4 6
5 10
6 15

Thanks for the help I appreciate it!
 
If there are 100 people and each person shakes hands only one time with each person how many handshakes would that be?

The answer is 4950. My question is how do I put this into an algebra formula?

# People # Handshakes

1 0
2 1
3 3
4 6
5 10
6 15

Person #1 shakes the hand of 99 other people.
Person #2 shakes the hand of 98 other people already having shaken the hand of person #1..
Person $3 shakes the hand of 97 other people already having shaken the hands of persons #1 and #2.

Do you see where this is heading?

The total number of handshakes becomes 99 + 98 + 97 + 96 +.....4 + 3 + 2 + 1.

Carl Freidrich Gauss had no idea as to how famous his name would become when he solved a seemingly time consuming arithmetic problem posed by his teacher with amazing speed. At age 10, his teacher presented his class with a long problem in addition, the answer to which he could find by formula in a few seconds. Some historical references indicate that the problem was of the following type, 72365 + 72550 + 72735 + .........+ 90865, where the common difference was 185, and the total number of terms was 100. Other sources claim that the class was asked to add up the first hundred whole numbers. As the story goes, it was customary at the school for the first person to get the answer to lay his paper or slate face down on the table. The teacher had barely finished presenting the problem when Gauss, alledgedly, placed his slate on the table, and sat quietly for the rest of the hour while his classmates struggled. Upon reviewing the slates at the end of the period, he found but a single number on Gauss's slate, the correct answer. Gauss supposedly did not know the trick for solving such arithmetic progression problems so quickly and his teacher was quite surprised.

History records that Gauss created the method instantaneously in his head. It is said that he did it by pairing the terms and then multiplying the value of each pair by the number of pairs. Taking the simpler problem of the first 100 integers to illustrate, the pairs all total 100, i.e., 100+0 = 100, 99+1 = 100, 98+2 = 100, .........51+49 = 100. This results in 50 pairs of 100 each totalling 5000, plus the 50 left over in the middle, for a total of 5050.

Other historical records say that he imagined the sum he sought, denoted by S, as being written in both ascending and descending order:
S = 1 + 2 + 3 + 4 + .....................................98 + 99 + 100
S = 100 + 99 + 98 + 97...........................................3 + 2 + 1

Now, instead of adding the numbers horizontally across the rows, Gauss supposedly added them vertically down the columns. In so doing, he derived S = 101 + 101 + 101 + 101 + ...............................101 + 101 + 101
as the sum of each column is simply 101. There being 100 columns, it was quickly obvious to Gauss that 2S = 100 x 101 = 10100 and therefore the sum of the first hundred whole numbers became S = 1 + 2 + 3 + 4 + .................+ 98 + 99 + 100 = 100(101)/2 = 10100/2 = 5050.

Considering the alledgedly more difficult problem posed by the teacher, and the fact that he supposedly wrote but a single number on his slate, I guess it would have been possible for Gauss to mentally add 72365 and 90865, multiply by 100, and divide by 2, giving him his single answer of 8,161,500. Obviously, the 1 through 100 version is a bit more believable. Either way, his teacher considered it a stroke of pure genius.

It is worth mentioning that any book on number theory will offer you the expressions for the sum of the first odd numbers 1 through n as (1+3+5+7+....+(2n-1) = n^2 and the sum of the first even numbers 2 through 2n as S(2+4+6+....(2n) = n(n-1). Therefore, if we add all the odd and even integers from 1 through n, we will get

S = (n/2)^2 + (n/2)(n/2 + 1) = n^2/4 + n^2/4 + n/2 = n^2/2 + n/2 = n(n + 1)/2

What do you know? Look familiar? S(1 + 100) = 100(101)/2 = 5050.

Applying this to your problem, the total number of handshakes would be S = 99(100)/2 = 4950.
 
Hello, kmring!

If there are 100 people and each person shakes hands only one time with each person,
how many handshakes would that be?

The answer is 4950. My question is how do I put this into an algebra formula?
Assuming you're new to this type of problem, here's a walk-through.

Take any person in the group, say, person \(\displaystyle A\).
He can shake hands with any of the other 99 people . . . so he creates 99 handshakes.

This is true for each of the 100 people, so there are: \(\displaystyle \,100\,\times\,99\:=\:9900\) handshakes.

But this list includes entries like: "A shakes hands with B" and "B shakes hands with A",
\(\displaystyle \;\;\)which represent the same handshake. \(\displaystyle \;\)So our list is twice as long as it should be.

Therefore, there are: \(\displaystyle \,\frac{100\,\times\,99}{2} \:=\:4950\) handshakes.


Generalization: For \(\displaystyle n\) people, there will be: \(\displaystyle \.\frac{n(n\,-\,1)}{2}\) handshakes.

You probably haven't learned Permutations and Combinations yet.
This is a Combination problem, written: \(\displaystyle \,_{_{100}}C_{_2}\,\) or \(\displaystyle \,C(100,2)\,\) or \(\displaystyle \frac{100!}{2!\cdot98!}\)
 
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