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- anonymous

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!

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- anonymous

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- anonymous

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..

- anonymous

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

- anonymous

Could you solve it using factorials, please?

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- anonymous

hmm i found the answer as 9900

- amistre64

if there are 3 people:
1,2 ; 1,3 ; 2,3 .... is that right?

- anonymous

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...

- anonymous

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

- amistre64

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

- anonymous

No, all the numbers from 1 to 99.

- amistre64

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

- amistre64

ack .... 99 then lol

- amistre64

wouldnt the last guy have noone to shake hands with?

- anonymous

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,

- anonymous

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.

- anonymous

yea you right

- amistre64

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!!

- anonymous

Glad we agree.

- amistre64

factorial would be:
100! 100*99
---------- = -------
2! (100-2)! 2

- anonymous

its the same with C(100,2)

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