2nc2 :nc2 = 12:1 then n=

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

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)!}

^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}.

19th February 2019

ncert solution

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