Question

100 people are present in a party. If each of them must do a handshake with all the other people, how many handshakes should be done? Please explain how to get the answer, thanks!

Answer 1

you need 2 people for an hand shake and u have 100 people so i think that its C(100,2) but i am not sue of it..

Answer 2

\[\sum_{n=1}^{99}n=\frac{1}{2}\times99\times100=4950\]

Answer 3

Could you solve it using factorials, please?

Answer 4

hmm i found the answer as 9900

Answer 5

if there are 3 people:

1,2 ; 1,3 ; 2,3 .... is that right?

Answer 6

Think about it like this: the first person has 99 people to shake hands with, the second person has 98 because they've already shaken hands with the first, the third has 97...

Answer 7

yea but its harder to calculate this like that... so i think C(100,2) better way of solve

Answer 8

5054 if we do that ..... add all the numbers from 1 to 100 :)

Answer 9

No, all the numbers from 1 to 99.

Answer 10

n(n+1)
------ :)
2

Answer 11

ack .... 99 then lol

Answer 12

wouldnt the last guy have noone to shake hands with?

Answer 13

mhmh yea but there is 100 people and they and we need 2 people to shake hands so that it should be 100 x 99 /2,

Answer 14

Yeah, the last guy has no-one to shake hands with, that's why it's sum of the first 99 integers, not the first 100.

Answer 15

yea you right

Answer 16

5.4321
4.321
3.21
2.1
1.0

i get 5 people with 10 shakes 5(4)/2 = 10

8.7654321
7.654321
6.54321
5.4321
4.321
3.21
2.1
1.0

8 people with 28 shakes 8(7)/2 = 28

I spose it works :)

100(99)/2 = 99*50 = 4950 yay!!

Answer 17

Glad we agree.

Answer 18

factorial would be:

100! 100*99
---------- = -------
2! (100-2)! 2

Answer 19

its the same with C(100,2)