binom{50}{4} + sum_{i=1}^{6} binom{56-i}{3} = | vedclass

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

Question

(D) We use the identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.Given expression: $\binom{50}{4} + \sum_{i=1}^{6} \binom{56-i}{3} = \binom{50

Vedclass · Accepted Answer

$\binom{56}{4}$

Vedclass · Answer

(D) We use the identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.Given expression: $\binom{50}{4} + \sum_{i=1}^{6} \binom{56-i}{3} = \binom{50}{4} + \binom{55}{3} + \binom{54}{3} + \binom{53}{3} + \binom{52}{3} + \binom{51}{3} + \binom{50}{3}$.Rearranging the terms:$= \binom{50}{4} + \binom{50}{3} + \binom{51}{3} + \binom{52}{3} + \binom{53}{3} + \binom{54}{3} + \binom{55}{3}$.Using $\binom{50}{4} + \binom{50}{3} = \binom{51}{4}$:$= \binom{51}{4} + \binom{51}{3} + \binom{52}{3} + \binom{53}{3} + \binom{54}{3} + \binom{55}{3}$.Using $\binom{51}{4} + \binom{51}{3} = \binom{52}{4}$:$= \binom{52}{4} + \binom{52}{3} + \binom{53}{3} + \binom{54}{3} + \binom{55}{3}$.Continuing this process:$= \binom{53}{4} + \binom{53}{3} + \binom{54}{3} + \binom{55}{3} = \binom{54}{4} + \binom{54}{3} + \binom{55}{3}$.$= \binom{55}{4} + \binom{55}{3} = \binom{56}{4}$.

$\binom{50}{4} + \sum_{i=1}^{6} \binom{56-i}{3} = \dots$

Similar Questions

If $^nC_r = C_r$ and $2 \frac{C_1}{C_0} + 4 \frac{C_2}{C_1} + 6 \frac{C_3}{C_2} + \dots + 2n \frac{C_n}{C_{n-1}} = 650$,then $^nC_2 =$

$\sum\limits_{r = 0}^m {^{n + r}{C_n} = } $

In the expansion of $(1 + x)^{50}$,the sum of the coefficients of odd powers of $x$ is

If ${ }^{n} C_0+\frac{1}{2}{ }^{n} C_1+\frac{1}{3}{ }^{n} C_2+\ldots+\frac{1}{n+1}{ }^{n} C_{n}=\frac{1023}{10}$,then $n=$

If $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + .... + a_{2n}x^{2n}$,then $a_0 + a_2 + a_4 + .... + a_{2n} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module