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.