binom{47}{4} + sum_{r=1}^5 binom{52-r}{3} = d | vedclass

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

Question

(C) We use the identity $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$.Given expression: $S = \binom{47}{4} + \sum_{r=1}^5 \binom{52-r}{3}$Expandi

Vedclass · Accepted Answer

$\binom{52}{4}$

Vedclass · Answer

(C) We use the identity $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$.Given expression: $S = \binom{47}{4} + \sum_{r=1}^5 \binom{52-r}{3}$Expanding the summation:$S = \binom{47}{4} + \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{47}{3}$Using $\binom{47}{4} + \binom{47}{3} = \binom{48}{4}$:$S = \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{48}{4} = \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{49}{4}$Using $\binom{49}{3} + \binom{49}{4} = \binom{50}{4}$:$S = \binom{51}{3} + \binom{50}{3} + \binom{50}{4} = \binom{51}{3} + \binom{51}{4}$Using $\binom{51}{3} + \binom{51}{4} = \binom{52}{4}$.Thus,the correct option is $C$.

$\binom{47}{4} + \sum_{r=1}^5 \binom{52-r}{3} = \dots$

Similar Questions

$\sum_{k=0}^{20} \left({}^{20}C_{k}\right)^{2}$ is equal to :

Choose the correct option regarding the following statements:
$1$. $C_0+C_2+C_4+\ldots+C_n=2^{n-1}$,if $n$ is even
$2$. $C_1+C_3+C_5+\ldots+C_{n-1}=2^{n-1}$,if $n$ is even

Let $m, n \in \mathbb{N}$ and $\operatorname{gcd}(2, n)=1$. If $30\binom{30}{0} + 29\binom{30}{1} + \ldots + 2\binom{30}{28} + 1\binom{30}{29} = n \cdot 2^m$,then $n + m$ is equal to (Here $\binom{n}{k} = {^nC_k}$)

If $\sum_{r=1}^{30} \frac{r^2({}^{30}C_r)^2}{{}^{30}C_{r-1}} = \alpha \times 2^{29}$,then $\alpha$ is equal to

Let $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. Then the sum $\frac{1}{2^{10}} \sum_{k=0}^{10} \binom{10}{k} k^2$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module