There are $3$ girls in a class of $10$ students. The number of different ways in which they can be seated in a row such that no two of the three girls are together is

  • A
    $7! \times ^6P_3$
  • B
    $7! \times ^8P_3$
  • C
    $7! \times 3!$
  • D
    $\frac{10!}{3!7!}$

Explore More

Similar Questions

Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals

If $T_n$ denotes the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides and $T_{n+1}-T_{n}=21$,then $n=$

Out of $30$ points in a plane,$8$ of them are collinear. The number of straight lines that can be formed by joining these points is:

Three parallel straight lines $L_1, L_2$ and $L_3$ lie on the same plane. Consider $5$ points on $L_1, 7$ points on $L_2$ and $9$ points on $L_3$. Then the maximum possible number of triangles formed with vertices at these points is:

Out of $18$ points in a plane,no three are in the same straight line except five points which are collinear. The number of $(i)$ straight lines and $(ii)$ triangles which can be formed by joining them is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo