Question

At a party, 55 handshakes took place as everyone shook hands with everyone else in attendance. How many people were at the party?

Answer 1

Suppose there were \(\rm \large n\) people at the party. It takes two persons for one handshake. How many different ways can you choose \(2\) people from the available \(\large \rm n\) people ?

Answer 2

If there are `n` people at the party, then each person will shake hands with `n-1` other people. So with n people each making (n-1) handshakes, it appears at first sight that there aren(n-1)handshakes.

However, each handshake will have been counted twice, i.e. A->B and
B->A, so we must divide by 2.

Total number of handshakes = \(\large \bf \frac{n(n-1)}{2}\)

Answer 3

and REMEMBER THIS FORMULA :-
\[\large \bf Total~No.Of~Handshakes=\frac{n(n-1)}{2}\]

Answer 4

@serjinio

Answer 5

so,we get
\[\large \bf \frac{n(n-1)}{2}=55\]
and solve for n ?

Answer 6

hope you understood ! @serjinio

Answer 7

lets say you have 4 people
A,B,C,D

the *different* handshakes are

AB, AC, AD
BC, BD
CD

Answer 8

handshake AB = BA

Answer 9

yes,so we divide by 2 because they are not 4 persons actually they are 2.
That's why, we divide it by 2
we get
No.of HANDSHAKES =n(n-1)/2

Answer 10

2 persons :- A and B