Question

if n!÷2! (n-2)! and n!÷4!(n-4)! are ratio 2:1 then find the value of n

Answer 1

n!/[2(n-2)!]:n!/[4(n-4)!]=2:1
[n!/2!(n-2)!].[n!/4!(n-4)!]=2/1
4.3.2!(n-4)!/[2!(n-2)(n-3)(n-4)!]=2/1
4.3/[(n-2)(n-3)]=2/1
(n-2)(n-3)=6
by solving we get n=0 or n=5

Answer 2

$\huge\underline\mathfrak\blue{Question:-}$

If $\sf\dfrac{n!}{2!(n-2)!}$ and $\sf\dfrac{n!}{4!(n-4)!}$

are in the ratio 2:1, find the value of n.

$\huge\underline\mathfrak\green{Answer:-}$

We have,

$\sf \qquad \: \frac{n!}{2!(n - 2)!} : \frac{n!}{4!(n - 4)!} = 2:1 \\ \\ \implies \sf \: \frac{n!}{2!(n - 2)!} \times \frac{4!(n - 4)!}{n!} = \frac{2}{1} \\ \\ \implies \sf \: \frac{4!(n - 4)!}{2!(n - 2) \times (n - 3) \times (n - 4)!} = \frac{2}{1} \\ \\ \implies \sf \: \frac{4 \times 3 \times 2!}{2!(n - 2)(n - 3)} = \frac{2}{1} \\ \\ \implies \sf \: (n - 2)(n - 3) = 6 \\ \\ \implies \sf \: (n - 2)(n - 3) = 3 \times 2 \\ \\ \implies \sf \: n - 2 = 3 \: \: \: and \: \: \: n - 3 = 2 \\ \\ \implies \sf \: n = 5$

Therefore, value of n will be 5.