MHB Solve the Handshake Problem: n Couples at a Party

[evinda]

Hello ! :)
Could you help me at the exercise below?
Suppose that n couples are at a party.
If every person at the party shake hands with any other person except from his partner, how many handshakes will have been exchanged?

 

[I like Serena]

Hello ! :)
Could you help me at the exercise below?
Suppose that n couples are at a party.
If every person at the party shake hands with any other person except from his partner, how many handshakes will have been exchanged?

Hi evinda!

Suppose we have 3 couples, say persons A, a, B, b, C, and c.
How many hands does A shake?
How many handshakes are there in total?
Can you generalize?

 

[evinda]

"A" and "a" shake 4 hands,"B" and "b" shake 2 hands.Can you give me a hint how to find the general formula,because I have stuck?

 

[I like Serena]

"A" and "a" shake 4 hands,"B" and "b" shake 2 hands.Can you give me a hint how to find the general formula,because I have stuck?

Actually, "A" and "a" shake 4 hands, "B" and "b" shake 4 hands, and "C" and "c" shake 4 hands.
So there are 6 x 4 times that someone shakes a hand.
Since it takes 2 persons to do a handshake, we should divide the total number by 2.
That means that the number of handshakes is 6 x 4 / 2 = 12.

Generalize?

 

[Evgeny.Makarov]

Here is an illustration.

 

[evinda]

Is it \frac{n\cdot (n-2)}{2} ,where n the number of persons that are at the party ?

 

[I like Serena]

Is it \frac{n\cdot (n-2)}{2} ,where n the number of persons that are at the party ?

Yep! ;)

Btw, in your problem statement, n was supposed to be the number of couples.
I'd advise against mixing up the meaning of symbols.
Your number of handshakes is \frac{2n\cdot (2n-2)}{2}, where $n$ is the number of couples.

 

[evinda]

Nice!Thank you very much! ;)