If {{n}!}/{2 ! n-2!} & {{n}!}/{4 !{n}-4!} are in the ratio 2: 1 then the value of {n} is

NIMCET 2021 Question Paper with solutions

Question : 3 of 120

Marks: +1, -0

{{n}!}/{2 ! n-2!} \ \ &\ {{n}!}/{4 !{n}-4!}

are in the ratio

2: 1

then the value of

{n}

Solution:

Given,

{{n}!}/{2 ! n-2!} : {{n}!}/{4 !{n}-4!}=2: 1

.......(1)
We can write above equation in the form

⇒{n!}{2! n-2!} \×{!4 n-4!}/{n!}={2}/{1}

\⇒{4.3}/{(n-2)(n-3)}={2}/{1}

\⇒(n-2)(n-3)=6

\⇒n^{2}-3 n-2 n+6=6

\⇒n^{2}-5 n=0

\⇒\ n(n-5)=0

\⇒n=0

n=5

But when we put

n=0

in equation (1)
It is not valid. Hence

n=5

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