The solution is for total 10 people and instead of 4 persons 3 are chosen
First find the total number of ways of selecting three people out of ten seated in a round table. Total number of selection is equal to10C3.
But the condition is that no two of them should be adjacent.
So first find the selections in which exactly two people are next to each other. Two people next to each other can be selected in10ways(AB,BC,CD,DE,EF,FG,GH,HI,IJ,JA). Once an adjacent pair of people selected, the remaining one person can be selected in 6 ways (so that he/she is not next to any of them).
So, total number of selections possible so that exactly two people are next to each other is equal to10×6=60
Also number ofselections in whichthree people are together is10.
(ABC, BCD........)
Hence, selections in which no two people are next to each other=total selection−( selections in which exactly two are together+selections in which all three are together)
=10C3−(60+10)=120−70=50
