The least value of the natural number n satis | vedclass

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

Question

(B) Using the Pascal's identity,${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$,we have ${}^nC_5 + {}^nC_6 = {}^{n+1}C_6$. Given inequality: ${}^{n+1}C_6 > {}^{

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) Using the Pascal's identity,${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$,we have ${}^nC_5 + {}^nC_6 = {}^{n+1}C_6$. Given inequality: ${}^{n+1}C_6 > {}^{n+1}C_5$. Expanding the combinations: $\frac{(n+1)!}{6!(n-5)!} > \frac{(n+1)!}{5!(n-4)!}$. Dividing both sides by $(n+1)!$ and simplifying: $\frac{1}{6!(n-5)!} > \frac{1}{5!(n-4)!}$. Since $6! = 6 	imes 5!$ and $(n-4)! = (n-4) 	imes (n-5)!$,we get: $\frac{1}{6 	imes 5!(n-5)!} > \frac{1}{5!(n-4)(n-5)!}$. Canceling common terms: $\frac{1}{6} > \frac{1}{n-4}$. This implies $n-4 > 6$,so $n > 10$. The least natural number $n$ satisfying this is $n = 11$.

The least value of the natural number $n$ satisfying $C(n, 5)+C(n, 6)>C(n+1, 5)$ is

Similar Questions

The product of $r$ consecutive natural numbers is always divisible by......

The number of $5$ card combinations out of a deck of $52$ cards such that there is exactly one ace in each combination 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.

The number of values of $r$ satisfying $^{69}C_{3r-1} - ^{69}C_{r^2} = ^{69}C_{r^2-1} - ^{69}C_{3r}$ is:

The number of integral solutions to the equation $x+y+z=21$,where $x \geq 1, y \geq 3, z \geq 4$,is equal to $..........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module