Party!

This can be solved using combinations where handshakes can be considered as a selection of 2 people where the order doesn't matter.

Therefore by writing the combination $nC_{2}$ in factorials form we have:

$\frac{n !}{2 ! . (n-2)!} = 66$

By multiplying both sides by [ 2! ] we get:

$\frac{n !}{ (n-2)!} = 132$

Then we can expand the numerator factorial so it becomes:

$\frac{n.(n-1).(n-2)!}{ (n-2)!} = 132$

by cancelling $(n-2)!$ and distributing $n$ on the bracket $(n-1)$ we get:

$n^2 - n - 132 = 0$

Factorizing this:

$(n+11)(n-12)$

Therefore n is -11 or 12, (-11 is refused because n cannot be a negative number)

Let's say there are n people at the party. The first person shakes hands n-1 times, since he does not shake hands with himself. The next person then shakes hands with n-2 people (since he has already shaken hands with the first person), and so on until the last person has already shaken everyone's hand by the time it's his turn.

The total number of shakes for a party with n people is then given by 1+2+3+...+(n-1).

The formula for summation of integers 1 through x is (x)(x+1)/2. The total number of shakes at a party with n people is therefore (n-1)(n)/2. If we say (n-1)(n)/2 = 66, then we can solve for n to find that n = 12.

66 "double" handshakes = 132 handshakes. 132 = 11 x 12, i.e. each of 12 people shakes hands with 11 other people.

lets say there are n persons first person shakes hand with everyone else: n-1 times(n-1 persons) second person shakes hand with everyone else(not with 1st as its already done): n-2 times 3rd person shakes hands with remaining persons: n-3 So total handshakes will be = (n-1) + (n-2) + (n-3) +…… 0; = (n-1) (n-1+1)/2 = (n-1) n/2 = 66 = n^2 -n = 132 =(n-12)(n+11) = 0; = n = 12 OR n =-11 -11 is ruled out so the answer is 12 persons.

you can use graph theory. A complete graph Kn(with n vertices) will have n(n-1))/2 edges. If we consider each handshake to be one edge, and n to be the number of people, then (n(n-1))/2=66 n(n-1)=66 x 2=132 n^2-n=132 n^2-n-132=0 (n-12)(n+11)=0 n=12 or n=-11(discard n=-11 as n >0) so there are 12 people at the party

Each person who arrives at the party shakes hands to everybody who arrived before; second shakes hands with first, thirth with first and second, and so on. Then you only have to add persons at first until the outcome is 66 handshakes. 1+2+3+4+5+6+7+8+9+10+11=66

Recursive solution

Let $h(n)$ be the total number of handshakes between $n$ persons, then:

For an empty party, or if there is only one person: $h(0)=h(1)=0$

If there are two persons: $h(2)=1$

For $n$ persons, person $n$ shakes his hand with $(n-1)$ persons: $h(n)=n-1$ , but we must consider handshakes between everyone not only between one person to the others, so we can call recursively for $(n-1)$ persons, so we get the recurrence:

$\left\{\begin{aligned} &h(0)=0\\ &h(n)=h(n-1)+(n-1) \end{aligned}\right.$

Find a closed form:

$\begin{aligned} h(n)&=h(n-1)+(n-1)\\ &=h(n-2)+(n-2)+(n-1)\\ &=h(0)+(1-1)+...+(n-2)+(n-1)\\ &=\sum_{k=1}^{n}(n-k)\\ &=\sum_{k=1}^{n}n-\sum_{k=1}^{n}k\\ &=n^2-\frac{n(n+1)}{2}\\ &=\frac{n(n-1)}{2}\\ \end{aligned}$

Then, for 66 handshakes:

$\frac{n(n-1)}{2}=66 \Rightarrow n^2-n-132 \Rightarrow \boxed{n=12}$

Let us say there are n people at the party, each person will shake the rest of (n-1) person. since A shakes B equals to B shakes A, (A B C: AB AC BA BC CA CB) so by n*(n-1) we have double counted the number of handshakes. Therefore n(n-1)/2 = 66, so n =12.

Amount of connections in the network of N subjects can be calculated as SQR(N) minus the summation of all numbers from 1 to N. You can use this formula to find the right answer as well

Just start counting your way to the top.

1+2+3+4=10

10+5+6=21

21+7+8+9=21+3×8=45

45+10+11=66

But you don't shake hands with yourself (unless you're lonely)

So we have to do 11+1=12 (too difficult?)

There is a further simple method to do this , just plug in the values in the formula : $\frac{n(n+1)}{2}$ where $n$ is the total number of hand shakes

solving this => $n^2 -n -132=0$

= $(n+11)(n-12)$

=> n=12 (ignoring the negative value since it is meaningless )

Rewriting the required as to find $n$ , where $\large ^{n}C_{2} = 66$ . $\large \therefore \frac{ ^{n}P_{2}}{2!} = 66$ $\large \therefore ^{n}P_{2} = 132 = 12 \times 11 = ^{12}P_{2}$ $\large \therefore n = \boxed{12}$ .

There is n people in the party. Everyone shakes hands with everyone ELSE, so everyone shakes hands with n - 1 people. There is n(n-1) handshakes, but this amount must be halved, because if person A shakes hands with person B it is same as if person B shakes hands with person A, so total amount of handshakes is n(n-1)/2. We know that total number of handshakes is 66, so n(n-1)/2 = 66 n^2 - n = 132 n^2 - n - 132 = 0 (n + 11)(n - 12) = 0 So solutions are -11 and -(-12)=12. Number of handshakes must be positive, so 12 is only solution left.

Result: 12

Here's the way that I did it, although there are more effective ways of doing it.

I figured out in my head that:

2 people = 1 hand shake 3 people = 3 hand shakes 4 people = 6 hand shakes 5 people = 10 hand shakes

Then I noticed the pattern and just kept adding up the pattern until I got to 66 handshakes.

66 can factorized as = 11 x 6 = 11 x 12 / 2 = 12 = handshake problem = n(n-1)