Mathematics - Probability Question with Solution | TestHub

Consider a group of three students $A,B$ and an other student in between $A$ and $B$. Choice for a student between $A$ and $B$ is 4. $A$ and $B$ can interchange their places in the group in 2 ways. Now the group of three students (student $A$, student $B$ and a student in between $A$ and $B$ ) and the remaining 3 students can be stand in a row in 4 ! ways. Hence total number of ways to stand in a row such that $A$ and $B$ are separated with one student in between them $=4\times 2\times 4$ ! Now total number of ways to stand 6 student stand in a row without any restriction $=6!$ Hence required probability $$=\frac{4\times 2\times 4!}{6!}=\frac{4\times 2}{6\times 5}=\frac{4}{15}$$