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

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(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

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(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

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