Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (hba):

Everybody in a room shakes hands with everybody else.the total number of handshakes is 66.Then the total number of persons in the room is :

OpenStudy (hba):

only hints

hartnn (hartnn):

for 2 ppl, handshakes = 1 for 3 people . handshakes = 1+2 for 4 people, handshakes = ?

OpenStudy (hba):

1+2+3

OpenStudy (hba):

@hartnn

OpenStudy (anonymous):

12 person in meeting

OpenStudy (hba):

I have reported you ^

hartnn (hartnn):

so continuing that way, could u generalize it ?

hartnn (hartnn):

for n ppl , 1+2+3+...n-1 = ?

OpenStudy (hba):

Ok so how does this help

hartnn (hartnn):

for n ppl, it will be n(n-1)/2 = 66 find n

OpenStudy (hba):

So where did you get n(n-1)/2 from ?

hartnn (hartnn):

1+2+3+4+.....n = sum of n numbers = n(n+1)/2 standard formula

OpenStudy (hba):

Thanks Alot.

OpenStudy (hba):

Here you have written n(n+1)/2 ?

hartnn (hartnn):

yes, for n for n-1 ?

hartnn (hartnn):

just replace n by n-1

OpenStudy (hba):

Ok so (n-1)(n+1)/2

hartnn (hartnn):

no

OpenStudy (anonymous):

suppose in a room n person are there.now two person make one hand shake.means total shake hand is C(n,2),but we are given total shake hand 66.solve C(n,2)=66 we have 12

hartnn (hartnn):

(n-1)(n-1+1)/2

OpenStudy (hba):

oh ok ok i got it now :) Thank you. let me find the answer.

OpenStudy (hba):

n(n-1)/2=66

OpenStudy (anonymous):

right

hartnn (hartnn):

ya, solve that

OpenStudy (anonymous):

12 right

OpenStudy (hba):

So i got n^2-n-132=0

hartnn (hartnn):

go on

OpenStudy (hba):

ok i am trying

OpenStudy (anonymous):

minus 12 and plus 11 solve this

OpenStudy (hba):

n=-11 and n=12

hartnn (hartnn):

number of pll can't be negative

OpenStudy (hba):

-11 cannot be the answer

OpenStudy (hba):

So 12 is the answer Thanks :)

hartnn (hartnn):

welcome :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!