Sixty Six

Let there be $n$ people at the party.

The first person must have shook hands with every body else. Thus he shook $(n-1)$ hands. The next person would have already shook hands with the first person So he will shake hands with whoever is left. Thus he will shake hands with $(n-2$ people. This would continue until there is one person left who would have shaken hands with every body.

So we have $(n-1)+(n-2)+(n-3).......2+1+0=66$

We know sum of $n$ terms : $\frac{n(n+1)}{2}$ thus here we will apply $\frac{(n-1)(n-1+1)}{2}$

We can guess that here $n$ would be somewhere around $9-14$ . On trying we get $\frac{11 \times 12}{2} = 66$

Thus, $n-1=11 \\ \rightarrow n=\boxed {12}$

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${ C }_{ 2 }^{ m }=66$ $\frac { m! }{ (m-2)!2! } =66$ $\frac { (m)(m-1)(m-2)! }{ (m-2)! } =132$ $m(m-1)=132$ ${ m }^{ 2 }-m-132=0$ $(m+11)(m-12)=0$ ${ m }_{ 1 }=-11,\quad { m }_{ 2 }=12$ Since -11 doesn't apply, the answer is 12

Note that handshakes will be repeated two times, and that each person shakes the number of total people at the party -1 times. You can work backwards by multiplying 66 by two, then find two sequential numbers whose product equals that.

66 X 2 = 132 ---> Known fact that 11 X 12 =132. Done

nc2=66. n=12

We can solve this by considering combinations.

We know that for a handshake, 2 people participate. So we have a total number of 66 ways for an X amount of people to shake hands, and each handshake represents a combination of 2 people. So basically, XC2 = 66.

By using the definition of nCr, 66 = X!/(2(X-2)!) 132 = X!/(X-2)! 132 = X(X-1) Which gives (X)^2-X-132 = 0

X = (1+sqrt(1+4*132))/2 = 12 As only the positive solution makes sense in this context, that of 12 people.

Well this can be solved using binomial terms, which can be expressed as C n,2 -> n!/(n-2)! 2!. So, this will be n . (n-1) .(n-2)!/(n-2)! .2

>> (n^2 - n)/2 = 66 You will get to a quadratic equation like this -> n^2 - n - 132=0 Solving in it you will get to 12 . No need of guessing!

i would rather go manually , 3 persons can shake hands in 3 ways , 4 persons in 3+(4-1) =6 ways, 5 persons in 6+(5-1)=10 ways............for 12 ,persons , 55+(12-1)=66

Using formula n(n+1)-2n/2 we get n=12

We can relate this to a $n$ sided polygon, where $n$ is the number of people. That is, $n$ vertices.

The total handshakes that occur are :

The number of diagonals $+ n$ {The adjacent points or vertices are not included in the diagonals, but the adjacent handshakes must be included}

$\Rightarrow \frac{n\cdot(n-3)}{2} + n = 66$ {As the number of diagonals in a n sided polygon is $\frac{n\cdot(n-3)}{2}$ }

$= n(n-3) + 2n = 66\times 2$

$= n^{2} - 3n + 2n - 132 = 0$

$= n^{2} - n -132 = 0$

$= n^{2} -12n + 11n -132 = 0$

$= n(n-12) + 11(n-12) = 0$

$= (n-12)\cdot(n+11) = 0$

Therefore $n=12$ or $n=-11$

But $-11$ people are not possible

Thus the number of people is ${n}$ , which is $\boxed{12}$ .

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.

Combinatory numbers, n elements making pairs (n 2) possibilities. (n 2)=n*(n-1)/2=66

Let n be the number of people. -nC2 -n!/(n-2)! (2!) -n(n-1)(n-2)(n-3).../(n-2)(n-3)..... (2!) -n(n-1)/2(1) -n(n-1)/2(1)=66 -n(n-1)=66*2 -n(n-1)=132 -12(11)=132 n=12

(N(n-3))/2 +n=66 N(n-3)+2n=132 N^2-3n+2n=132 N^2-n=132 N^2-n-132=0 (N+11)(n-12)=0 N=-11,12 N=12