# Easy Permutation-Combination Solved QuestionAptitude Discussion

 Q. After every get-together every person present shakes the hand of every other person. If there were $105$ handshakes in all, how many persons were present in the party?
 ✖ A. 14 ✖ B. 13 ✔ C. 15 ✖ D. 16

Solution:
Option(C) is correct

Let total number of persons present in the party be $x$,

Then,

$\dfrac{x(x-1)}{2} = 105$

$x = \textbf{15}$

## (2) Comment(s)

Raj Singh
()

2 people can have 1 hand shake

3 people can have 2+1 hand shake

4 people can have 3+2+1 hand shake

15 people can have 14+13+12+....+1 hand shake = 105 = n(n+1)/2

RAKESH
()

Total number of shake hand is 105.

every person shake hand with each other so that here must use combination

nC2=105

n!/(n-2)!2!=105

n.(n-1)(n-2)!/(n-2)!=210

n^2-n-210=0

(n-15)(n+14)=0

n=15 or n=-14

so the ans is 15.