Maths Handshake Puzzle

Anyone can solve the problem just by looking at the $CHOICES$ .

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

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

$n \times (n-1)=132$

$n^2 -n-132=0$

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

$n_1=12$ or $n_2 = -11$

Hence, the answer is $\boxed {12}$

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
...........ans-12

nC2=66 => n=12.

1. 1+2+3+4+5 ... +11 add them until you find 66. For 1 two people shakes hand so for 11it is twelve .
2. 1+2+3+4+...+11 sum(66) = (n/2)[2a+(n-1)d] //Sum of A.P a=1, d=1 Ans. n=11.

ı didn’t even do any mathematics — 12 is the only reasonable choice there.