The party mess

This can also be solved using the formula for combinations: ${ _{ n }{ C }_{ r }=\frac { n! }{ r!\left( n-r \right) ! } }$ Replacing r with 2, we get ${ _{ n }{ C }_{ 2 }=\frac { n! }{ 2!\left( n-2 \right) ! } \\ _{ n }{ C }_{ 2 }=\frac { n! }{ 2\left( n-2 \right) ! } \\ 2_{ n }{ C }_{ 2 }=\frac { n! }{ \left( n-2 \right) ! } \\ 2_{ n }{ C }_{ 2 }=\frac { n\cdot \left( n-1 \right) \cdot (n-2)! }{ \left( n-2 \right) ! } \\ 2_{ n }{ C }_{ 2 }=n\cdot \left( n-1 \right) \\ 2_{ n }{ C }_{ 2 }={ n }^{ 2 }-n }$ Since the number of handshakes is the same as the number of unique pairings (combinations of 2), ${ 2\left( 66 \right) ={ n }^{ 2 }-n\\ 132={ n }^{ 2 }-n\\ 0={ n }^{ 2 }-n-132\\ 0=\left( n-12 \right) \left( n+11 \right) \\ \\ \begin{matrix} n-12=0; & n+11=0 \\ n=12; & n=-11 \end{matrix} }$ Discarding the negative value, we now get 12 as the answer.

Solution: 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.

Total number of shake hands = n(n – 1)/2 = 66

n(n – 1) = 132 = 11 × 12

n = 11