There are $15$ stations on a train route and the train has to be stopped at exactly $5$ stations among these $15$ stations. If it stops at at least two consecutive stations,then the number of ways in which the train can be stopped is

  • A
    ${ }^{11} C_5$
  • B
    ${ }^{15} C_5$
  • C
    ${ }^{15} C_5 - { }^{11} C_5$
  • D
    ${ }^{15} C_{10} - { }^9 C_5$

Explore More

Similar Questions

The digit in the unit place of the number $(183!) + (3^{183})$ is

The number of permutations of all the letters $AAAABBBC$ in which all the $A$'s appear together in a block of $4$ letters or all the $B$'s appear together in a block of $3$ letters,is-

Suppose $a_2, a_3, a_4, a_5, a_6, a_7$ are integers such that $\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} + \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!}$,where $0 \leq a_j < j$ for $j = 2, 3, 4, 5, 6, 7$. The sum $a_2 + a_3 + a_4 + a_5 + a_6 + a_7$ is:

The remainder obtained when $1! + 2! + 3! + \dots + 11!$ is divided by $12$ is

The value of $(2 \cdot {}^{1}P_{0} - 3 \cdot {}^{2}P_{1} + 4 \cdot {}^{3}P_{2} - \dots \text{ up to } 51^{\text{th}} \text{ term}) + (1! - 2! + 3! - \dots \text{ up to } 51^{\text{th}} \text{ term})$ is equal to

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