MCQ
If six students, including two particular students $A$ and $B,$ stand in a row, then the probability that $A$ and $B$ are separated with one student in between them is
- A$\frac {8}{15}$
- ✓$\frac {4}{15}$
- C$\frac {2}{15}$
- D$\frac {1}{15}$
✓
Answer
Correct option: B.
$\frac {4}{15}$
b
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}}$
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}}$
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