Permutation-Combination
Aptitude

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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

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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.