1 votes 1 votes Prove that in a group of 6 people, there are at least three mutual friends or three mutual strangers. Rajesh Pradhan asked Feb 22, 2016 Rajesh Pradhan 2.3k views answer comment Share Follow See 1 comment See all 1 1 comment reply Bhagirathi commented Feb 25, 2016 reply Follow Share https://m.facebook.com/groups/1675844779330612?notif_t=group_activity&ref=m_notif 0 votes 0 votes Please log in or register to add a comment.
Best answer 5 votes 5 votes Hi. i think pigeonhole principle will be applied here.. two possibilities=> i) Mutual Friend,ii)stranger (let K be the number of possibilities K=2 here) now 6 people is there so N=6. ceil(N/K)=ceil(6/2)=3 atleast 3 of them will have to be either friend or stranger This is the approach Aboveallplayer answered Feb 22, 2016 • selected Mar 14, 2016 by Pooja Palod Aboveallplayer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes Use pigeon hole principal : There are 2 types of people either friends or strangers . Therefore $\left \lceil N/2 \right \rceil$ = 3 . Hence there would be either three mutual friends or 3 strangers . Riya Roy(Arayana) answered Feb 22, 2016 Riya Roy(Arayana) comment Share Follow See all 0 reply Please log in or register to add a comment.