1. Hand shake????

At a party, everyone shook hands with everybody else.
There were 66 handshakes. How many people were at the
party?

2. Re: Hand shake????

Originally Posted by abhforum
At a party, everyone shook hands with everybody else.
There were 66 handshakes. How many people were at the
party?
Can you solve $\dbinom{n}{2}=66~?$

Thanks a lot

4. Re: Hand shake????

basic drawn out method

Triangle numbers . . .

1 person shake hands 0 times (set n=0 as 1 person cannot create a handshake)
2 people shake hands 1 time set n=1
3 people shake hands 3 times set n=2
4 people shake hands 6 times set n=3 etc

n(n+1) = 66
2

n(n+1) = 132

therefore you want two numbers with a difference of 1 that multiply together to give you 132.

11x12=132 therefore n=11

5. Re: Hand shake????

A slightly different way of looking at it: each of the n people shakes hands with each of the n- 1 hand shakes. So n people have n(n- 1) hand shakes. But each hand shake involves 2 people so the actual number of hand shakes is n(n- 1)/2. Solve n(n-1)/2= 66

6. Re: Hand shake????

Originally Posted by mathmonster
basic drawn out method

Triangle numbers . . .

1 person shake hands 0 times (set n=0 as 1 person cannot create a handshake)
2 people shake hands 1 time set n=1
3 people shake hands 3 times set n=2
4 people shake hands 6 times set n=3 etc

n(n+1) = 66
2

n(n+1) = 132

therefore you want two numbers with a difference of 1 that multiply together to give you 132.

11x12=132 therefore n=11
You are off a little. It is (n-1)(n)/2=66 or (n-1)(n)= 132=11*12. So n=12