If frac{{}^{n+2} C_6}{{}^{n-2} C_2} = 11,then | vedclass

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

Question

(C) Given the equation $\frac{{}^{n+2} C_6}{{}^{n-2} C_2} = 11$. Using the formula ${}^{n} C_r = \frac{n!}{r!(n-r)!}$,we have: $\frac{(n+2)!}{6!(n-4)!

Vedclass · Accepted Answer

$n^2+3n-108=0$

Vedclass · Answer

(C) Given the equation $\frac{{}^{n+2} C_6}{{}^{n-2} C_2} = 11$. Using the formula ${}^{n} C_r = \frac{n!}{r!(n-r)!}$,we have: $\frac{(n+2)!}{6!(n-4)!} 	imes \frac{2!(n-4)!}{(n-2)!} = 11$ $\frac{(n+2)(n+1)n(n-1)}{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1} 	imes \frac{2 	imes 1}{1} = 11$ $\frac{(n+2)(n+1)n(n-1)}{360} 	imes 2 = 11$ $(n+2)(n+1)n(n-1) = 11 	imes 180 = 1980$ Testing values for $n$: If $n=9$,$(11)(10)(9)(8) = 7920 
eq 1980$. Wait,let us re-evaluate the expression: $\frac{(n+2)(n+1)n(n-1)}{6 	imes 5 	imes 4 	imes 3} = 11$ $(n+2)(n+1)n(n-1) = 11 	imes 360 = 3960$. For $n=9$,$11 	imes 10 	imes 9 	imes 8 = 7920$. Let us check the quadratic $n^2+3n-108=0$: $(n+12)(n-9)=0$,so $n=9$. Substituting $n=9$ into the original expression: $\frac{{}^{11} C_6}{{}^{7} C_2} = \frac{462}{21} = 22$. There seems to be a typo in the problem statement provided. Given the options,$n=9$ is the root for $n^2+3n-108=0$.

If $\frac{{}^{n+2} C_6}{{}^{n-2} C_2} = 11$,then $n$ satisfies the equation

Similar Questions

An aqueous solution of potash alum gives:

The time constant of an inductance coil is $3 \text{ ms}$. When a $90 \Omega$ resistance is joined in series,the time constant becomes $0.5 \text{ ms}$. The inductance and the resistance of the coil are:

Phytochrome is involved in:

Let $\alpha \neq 1$ be a real root of the equation $x^3-ax^2+ax-1=0$,where $a \neq -1$ is a real number. Then,a root of this equation,among the following,is

The wavelength of $H_{\alpha}$ line of Balmer series is $X \ \mathring{A}$. What is the wavelength of $H_{\beta}$ line of Balmer series?

Vedclass Test Series

Exam Paper Generator

Online Exam Module