The value of {}^{47}C_4 + sum_{j=1}^5 {}^{(52 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We use the identity ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$. Expanding the summation: $S = {}^{47}C_4 + ({}^{51}C_3 + {}^{50}C_3 + {}^{49}C_3 + {}^{

Vedclass · Accepted Answer

${}^{52}C_4$

Vedclass · Answer

(A) We use the identity ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$. Expanding the summation: $S = {}^{47}C_4 + ({}^{51}C_3 + {}^{50}C_3 + {}^{49}C_3 + {}^{48}C_3 + {}^{47}C_3)$. Rearranging terms: $S = ({}^{47}C_4 + {}^{47}C_3) + {}^{48}C_3 + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3$. Using the identity ${}^{47}C_4 + {}^{47}C_3 = {}^{48}C_4$: $S = ({}^{48}C_4 + {}^{48}C_3) + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3$. Using the identity ${}^{48}C_4 + {}^{48}C_3 = {}^{49}C_4$: $S = ({}^{49}C_4 + {}^{49}C_3) + {}^{50}C_3 + {}^{51}C_3$. Continuing this process: $S = ({}^{50}C_4 + {}^{50}C_3) + {}^{51}C_3 = {}^{51}C_4 + {}^{51}C_3 = {}^{52}C_4$. Thus,the correct option is $A$.

The value of ${}^{47}C_4 + \sum_{j=1}^5 {}^{(52-j)}C_3$ is

Similar Questions

The smallest positive integer $n$,for which $n! < {\left( {\frac{{n + 1}}{2}} \right)^n}$ holds is

$A$ pack contains $n$ cards numbered from $1$ to $n$. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is $1224$. If the smaller of the numbers on the removed cards is $k$,then $k - 20 =$

The value of $\sum\limits_{r = 1}^{15} {{r^2} \left( \frac{^{15}C_r}{^{15}C_{r - 1}} \right)}$ is equal to

If $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$,then $\sum_{r=0}^n \frac{r}{^nC_r}$ equals

$2^n \{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 3) \cdot (2n - 1)\} = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module