## anonymous 5 years ago 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!

1. 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..

2. anonymous

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

3. anonymous

Could you solve it using factorials, please?

4. anonymous

hmm i found the answer as 9900

5. amistre64

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

6. 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...

7. anonymous

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

8. amistre64

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

9. anonymous

No, all the numbers from 1 to 99.

10. amistre64

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

11. amistre64

ack .... 99 then lol

12. amistre64

wouldnt the last guy have noone to shake hands with?

13. 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,

14. 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.

15. anonymous

yea you right

16. 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!!

17. anonymous

18. amistre64

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

19. anonymous

its the same with C(100,2)