Handshakes: write the formula for number of handshakes that

[tomtom]

How would I write the formula for number of handshakes that 5, 6, 7, or 8 people can shake without shaking the same personj's hand twice?

If 5 People than 10 Handshakes,
6 People than 15 Handshakes,
7 People than 21 Handshakes,
8 People than 28 Handshakes.

H = Handshakes
P = People

I started with "H= [H-(P-1)]-(P-1)"....???, then I'm blanking out

Thank You.

 

[arthur ohlsten]

let th five people be ABCD and E
A can shake hands with 4 others
B can shake hands with 3 others[not with A ,already counted]
C can shake hands with 2 others [ not with A or B]
D can shake hands with 1 other [not A,B or C]

5 people can shake hands with 4*3*2*1
this is 4!

n people can shake hands for a total of [n-1]!

Arthur

 

[tomtom]

Perhaps I should've asked this way.... What is the formula for n if n is number of people present if there was let's say 21 Handshake in the meeting room and again one person only shook hands with others in the room only once with each and every other one?


(Correct numbers are below)
If 5 People than 10 Handshakes,
6 People than 15 Handshakes,
7 People than 21 Handshakes,
8 People than 28 Handshakes.

Thanks Again in advance.....

 

[Mrspi]

Perhaps I should've asked this way.... What is the formula for n if n is number of people present if there was let's say 21 Handshake in the meeting room and again one person only shook hands with others in the room only once with each and every other one?


(Correct numbers are below)
If 5 People than 10 Handshakes,
6 People than 15 Handshakes,
7 People than 21 Handshakes,
8 People than 28 Handshakes.

Thanks Again in advance.....

There are n people in the room. If each person shakes hands with everyone else in the room, then each person shakes hands with (n - 1) others. So, it might seems like there are n*(n - 1) handshakes. HOWEVER, this counts each handshake twice, since person A shaking hands with person B is really the same handshake as when person B shakes hands with person A.

So, the actual number of different handshakes is (1/2)*n*(n - 1).

handshakes = (1/2)*n*(n - 1)

I hope this is what you are looking for.

 

[Denis]

total handshakes h = 1 + 2 + 3 + ... + (n-1) where n = number of people

h = n(n-1) / 2

n = [sqrt(8h + 1) - 1] / 2

 

[arthur ohlsten]

sorry I goofed.
its 4+3+2+1=10

let S[n-1]= 1+2+3+...[n-1]
S[n-1]= 1+[n-1] times [n-1]/2
S[n-1]=n[n-1] / 2

or for n people number of handshakes= n[n-1]/2

Arthur

 

[soroban]

Re: Handshakes

Hello, tomtom!

Write the formula for number of handshakes that 5,6,7, or 8 people
can shake without shaking the same person twice?

5 People \(\displaystyle \:\rightarrow\:\) 10 Handshakes
6 People \(\displaystyle \:\rightarrow\:\) 15 Handshakes
7 People \(\displaystyle \:\rightarrow\:\) 21 Handshakes
8 People \(\displaystyle \:\rightarrow\:\) 28 Handshakes

Note the source of the numbers . . .

. . \(\displaystyle \begin{array}{ccc}\frac{5\cdot4}{2}\:=\:10 \\ \\\frac{6\cdot5}{2}\:=\:15 \\ \\ \frac{7\cdot6}{2}\:=\:21 \\ \\ \frac{8\cdot7}{2}\:=\:28 \end{array}\)


Therefore: \(\displaystyle \,n\) People \(\displaystyle \:\rightarrow\;\frac{n(n\,-\,1)}{2}\) Handshakes

 

[tomtom]

WoW!

Thanks EVERYBODY!!!!!

This was EXACTLY What I was looking for!!!!!

You guys (gals too) are GREAT!

THANKS x100 !!!!

-Tom (aka TomTom)

 

[tomtom]

total handshakes h = 1 + 2 + 3 + ... + (n-1) where n = number of people

h = n(n-1) / 2

n = [sqrt(8h + 1) - 1] / 2

Denis,
The first one works great!

But would you double check for me and see if

n = [sqrt(8h + 1) - 1] / 2

works or

n = [sqrt(8h + 1) + 1] / 2

works?


Thank You.