$\sum \limits_{k=0}^{6} {}^{51-k}C_{3}$ is equal to

  • A
    ${}^{51}C_{4}-{}^{45}C_{4}$
  • B
    ${}^{51}C_{3}-{}^{45}C_{3}$
  • C
    ${}^{52}C_{4}-{}^{45}C_{4}$
  • D
    ${}^{52}C_{3}-{}^{45}C_{3}$

Explore More

Similar Questions

For $r=0, 1, \ldots, 10$,let $A_{r}, B_{r}$ and $C_{r}$ denote,respectively,the coefficient of $x^{r}$ in the expansions of $(1+x)^{10}$,$(1+x)^{20}$ and $(1+x)^{30}$. Then $\sum_{r=1}^{10} A_r(B_{10} B_r - C_{10} A_r)$ is equal to

The value of $\frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \dots$ is equal to

Difficult
View Solution

If $26 \left( \frac{2}{3} \binom{12}{2} + \frac{2}{5} \binom{12}{4} + \frac{2}{7} \binom{12}{6} + \dots + \frac{2}{13} \binom{12}{12} \right) = 3^{13} - \alpha$, then $\alpha$ is equal to:

$2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + 11 \cdot {}^{20}C_3 + \dots + 62 \cdot {}^{20}C_{20}$ is equal to

If $n$ is a positive integer, the value of $(2n+1) ^nC_0 + (2n-1) ^nC_1 + (2n-3) ^nC_2 + \ldots + 1 \cdot ^nC_n$ 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