The coefficient of the middle term in the expansion of $(1 + x)^{10}$ is

  • A
    $\frac{10!}{5! \times 6!}$
  • B
    $\frac{10!}{(5!)^2}$
  • C
    $\frac{10!}{5! \times 7!}$
  • D
    None of these

Explore More

Similar Questions

Show that the middle term in the expansion of $(1+x)^{2n}$ is $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} 2^n x^n$,where $n$ is a positive integer.

If the coefficients of $r$ th and $(r+1)$ th terms in the expansion of $(3+7x)^{29}$ are equal,then $r$ is equal to

If the coefficients of $x^9$ and $x^{10}$ in the binomial expansion of $(3+\frac{x}{2})^n$ are equal,then $n=$

What is the coefficient of $x^{100}$ in $(1 + x + x^2 + x^3 + \dots + x^{100})^3$?

Let the coefficients of $x^{-1}$ and $x^{-3}$ in the expansion of $(2x^{1/5} - x^{-1/5})^{15}$,$x > 0$,be $m$ and $n$ respectively. If $r$ is a positive integer such that $mn^2 = {}^{15}C_r \cdot 2^r$,then the value of $r$ 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