Question

2nc2 :nc2 = 12:1 then n=?

Answer 1

Since it is given that

$^2^n C_{2}:^nC_{2}=12:1$

By using the combination formula which states:

$^nC_{r}=\frac{n!}{r!(n-r)!}$

Now, $^2^n C_{2}:^nC_{2}=12:1$

${\frac{(2n)!}{2!(2n-2)!}} \times \frac{(n-2)!2!}{n!}= 12$

${\frac{(2n)(2n-1)(2n-2)!}{(2n-2)!}} \times \frac{(n-2)!}{n(n-1)(n-2)!}= 12$

$\frac{(2n)(2n-1)}{n(n-1)}= 12$

$\frac{2(2n-1)}{(n-1)}= 12$

$(2n-1)= 6(n-1)$

2n-1 = 6n-6

2n-6n = -6+1

-4n = -5

n = $\frac{5}{4}$

So, the value of 'n' is $\frac{5}{4}$.

Answer 2

Answer: Can not be determined.

Step-by-step explanation:

$\frac{^{2n}C_2}{^nC_2}=\frac{12}{1}$

$\implies \frac{(2n)!/2!(2n-2)!}{n!/2!(n-2)!}=\frac{12}{1}$

$\implies \frac{2n(2n-1)}{n(n-1)}=\frac{12}{1}$

$\implies 2n (2n-1)=12n(n-1)$

$\implies 2n-1 = 6(n-1)$

$\implies -4n = -5\implies n = \frac{5}{4}$

Thus, the value of n is 5/4

Which is not possible.

Hence, the value of n can not be determined with the given expression.