The number of values of n in N for which { }^ | vedclass

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

Question

(A) Given the ratio: $\frac{{ }^{n+2} C_2}{{ }^{n+3} C_1} = \frac{4}{2} = 2$. Using the formula ${ }^n C_r = \frac{n!}{r!(n-r)!}$,we have: $\frac{(n+2

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given the ratio: $\frac{{ }^{n+2} C_2}{{ }^{n+3} C_1} = \frac{4}{2} = 2$. Using the formula ${ }^n C_r = \frac{n!}{r!(n-r)!}$,we have: $\frac{(n+2)!}{2! n!} \div (n+3) = 2$. $\frac{(n+2)(n+1)}{2} 	imes \frac{1}{n+3} = 2$. $(n+2)(n+1) = 4(n+3)$. $n^2 + 3n + 2 = 4n + 12$. $n^2 - n - 10 = 0$. Solving for $n$ using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $n = \frac{1 \pm \sqrt{1 - 4(1)(-10)}}{2} = \frac{1 \pm \sqrt{41}}{2}$. Since $\sqrt{41}$ is not an integer,$n$ cannot be a natural number $(n 
otin N)$. Therefore,the number of values of $n$ is $0$.

The number of values of $n \in N$ for which ${ }^{n+2} C_2 : { }^{n+3} C_1 = 4 : 2$ is

Similar Questions

Let $S = \{1, 2, 3, \ldots, 9\}$. For $k = 1, 2, \ldots, 5$,let $N_k$ be the number of subsets of $S$,each containing five elements out of which exactly $k$ are odd. Then $N_1 + N_2 + N_3 + N_4 + N_5 =$

If ${}^{10}P_r = 604800$ and ${}^{10}C_r = 120$,then $r =$

The number of ways in which $17$ apples can be distributed among $4$ guests such that each guest gets at least $3$ apples is

In a library,there are $n$ distinct books and each has $p$ copies. What is the number of ways to select one or more books?

The greatest integer which divides $(p+1)(p+2)(p+3) \ldots (p+q)$ for all $p \in N$ and fixed $q \in N$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module