Handshake Puzzle

$_nC_2=\frac {n!} {2!(n-2)!}$

$_nC_2=\frac {n(n-1)} {2!}$

$n^2-n=2!(66)$

$n^2-n-132=0$

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

$n=12, -11$ but n can't be negative

Or you may solve it this way: The first person would be making $n-1$ handshakes as he won't with himself. The second person would make $n-2$ as he wont shake hands again with the first and to himself. The third would make $n-3$ , the fourth $n-4$ and so on till there are only 2 left to make 1 handshake. If you add all of these we get $1+2+3+...+(n-1)=\frac {(n-1)(n)} {2}=66$

The first person will shake hands with everyone else (n-1).

The second will shake hands with all except the first person (n-2), and so on.

The simple way to look at it is that it will be sum of the integers up to (n-1), i.e. a triangle number. 66 is the 11th triangle number, so the number of people in the room is 12.

A more mathematical way would be to use the formula to sum the terms: n(n-1)/2 = 66.

This can be seen from adding the first and the last term together (n-1) + 1 = n. If we add the second and second to last term we also get n, and so on. We can do this half the number of total terms: (n-1)/2, and thus the equation which solves to 12.

Number of handshakes $=\frac{n(n-1)}{2}=66$

$n^{2}-n=132$

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

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

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

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

So, values of $n= 12$ and $-11$

But, $n$ cannot be negative.

Thus, the answer is: $n=\boxed{12}$

$_nC_2=\frac {n!} {2!(n-2)!}$

$_nC_2=\frac {n(n-1)} {2!}$

$n^2-n=2!(66)$

$n^2-n=132$

$n^2-n-132=0$

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

$n=12, -11$

Here n cannot be negative as well as we all know a negative quantity doesn't exist so $n=12$

We can think of handshakes as a relation R between people on a set S, where aRb (a has shaken hands with b) and bRa exist for all a and b in S, and there exists no element xRx for all x in S. In other words, you can't shake your own hand.

With that all in mind, we can calculate the number of elements are in our relation, assuming S is order n as: n^2 - n. (number of elements in our relation assuming every relation between elements of S is possible, minus the only ones that aren't possible in our relation, the n reflexive relations from an x to itself -- shaking your own hand.)

This then leaves us with twice the number of handshakes we need, since we aren't counting aRb and bRa as distinct relationships, so:

(n^2-n)/2, or factored out into the more familiar formula: n(n-1)/2 is the number of handshakes occur if n people have shaken hands with everyone else.

Now, solving for n in terms of h:

n = 1/2(sqrt(8h+1)+1)

substituting in n=66 then gives 12.

12

In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n. Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 66. This is the quadratic equation n2+ n -132 = 0. Solving for n, we obtain 11 as the answer and deduce that there were 12 people at the party.

Since 66 is a relatively small number, you can also solve this problem with a hand calculator. Add 1 + 2 = + 3 = +... etc. until the total is 66. The last number that you entered (11) is n.