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NIMCET 2021 Question Paper with solutions

Question : 3 of 120

Marks: +1, -0

\frac{n!}{2!\,(n-2)!} \qquad \frac{n!}{4!\,(n-4)!}

2:1

n

Solution:

Given,

\frac{n!}{2!\,(n-2)!} : \frac{n!}{4!\,(n-4)!}=2:1

.......(1)

We can write above equation in the form

\Rightarrow \frac{n!}{2!\,(n-2)!} \times \frac{4!\,(n-4)!}{n!} = \frac{2}{1}

\Rightarrow \frac{4.3}{(n-2)(n-3)} = \frac{2}{1}

\Rightarrow (n-2)(n-3)=6

\Rightarrow n^{2}-3n-2n+6=6

\Rightarrow n^{2}-5n=0

\Rightarrow n(n-5)=0

\Rightarrow n=0

n=5

But when we put

n=0

in equation (1)

It is not valid. Hence

n=5

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