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

If $4$ red balls and $5$ green balls are selected from $n$ balls,and the sum of both selections is greater than ${}^{n+1}C_4$,then the value of $n$ is equal to:

$A$ team of $5$ students is to be selected from $12$ students. If two particular students are to be included in that team,then the number of ways that such team can be selected is

There are two urns. Urn $A$ contains $3$ distinct red balls,and Urn $B$ contains $9$ blue balls. If two balls are selected at random from each urn and swapped,in how many ways can this operation be performed?

The number of ways to select one or more books from $6$ books is:

The domain of definition of the function $f(x) = \log_{[x + \frac{1}{x}]} |x^2 - x - 6| + ^{16-x}C_{2x-1} + ^{20-3x}P_{2x-5}$ is,where $[x]$ denotes the greatest integer function.

Vedclass Test Series

Exam Paper Generator

Online Exam Module