The value of {}^{50}C_4 + sum_{r=1}^{6} {}^{5 | vedclass

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

Question

(B) We are given the expression $S = {}^{50}C_4 + \sum_{r=1}^{6} {}^{56-r}C_3$. Expanding the summation,we get $S = {}^{50}C_4 + ({}^{55}C_3 + {}^{54}

Vedclass · Accepted Answer

${}^{56}C_4$

Vedclass · Answer

(B) We are given the expression $S = {}^{50}C_4 + \sum_{r=1}^{6} {}^{56-r}C_3$. Expanding the summation,we get $S = {}^{50}C_4 + ({}^{55}C_3 + {}^{54}C_3 + {}^{53}C_3 + {}^{52}C_3 + {}^{51}C_3 + {}^{50}C_3)$. Rearranging the terms,$S = ({}^{50}C_4 + {}^{50}C_3) + {}^{51}C_3 + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3$. Using the Pascal's identity ${}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r$,we have ${}^{50}C_4 + {}^{50}C_3 = {}^{51}C_4$. Now,$S = ({}^{51}C_4 + {}^{51}C_3) + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3 = {}^{52}C_4 + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3$. Continuing this process,$S = ({}^{52}C_4 + {}^{52}C_3) + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3 = {}^{53}C_4 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3$. $S = ({}^{53}C_4 + {}^{53}C_3) + {}^{54}C_3 + {}^{55}C_3 = {}^{54}C_4 + {}^{54}C_3 + {}^{55}C_3$. $S = ({}^{54}C_4 + {}^{54}C_3) + {}^{55}C_3 = {}^{55}C_4 + {}^{55}C_3 = {}^{56}C_4$. Thus,the correct option is $B$.

The value of ${}^{50}C_4 + \sum_{r=1}^{6} {}^{56-r}C_3$ is

Similar Questions

There are $25$ candidates for $12$ positions,out of which $5$ are from the reserved category. If $3$ positions are reserved and the remaining are open,in how many ways can the selection be made?

If $\frac{1}{{ }^{5}C_{r}} + \frac{1}{{ }^{6}C_{r}} = \frac{1}{{ }^{4}C_{r}},$ then the value of $r$ is

There are $5$ types of ice creams available in a shop. $A$ child buys $6$ ice creams.
Statement-$1$: The child can buy $6$ ice creams in $^{10}C_5$ ways.
Statement-$2$: The number of ways the child can buy $6$ ice creams is equal to the number of ways to arrange $6$ '$A$'s and $4$ '$B$'s in a row.

$A$ class contains $b$ boys and $g$ girls. If the number of ways of selecting $3$ boys and $2$ girls from the class is $168$,then $b + 3g$ is equal to.

The number of groups that can be made from $5$ different green balls,$4$ different blue balls,and $3$ different red balls,if at least $1$ green and $1$ blue ball must be included.

Vedclass Test Series

Exam Paper Generator

Online Exam Module