If $\frac{\mathrm{n} !}{(2 !) \times(\mathrm{n}-2) !}: \frac{\mathrm{n} !}{(4 !) \times(\mathrm{n}-4) !}=2: 1$, find the value of $\mathrm{n}$
Given Equation :
$\frac{n !}{(2 !) \times(n-2) !}: \frac{n !}{(4 !) \times(n-4) !}=2: 1$
To Find : Value of n
Formula $n !=n \times(n-1) !$
By given equation,
$\frac{n !}{(2 !) \times(n-2) !}: \frac{n !}{(4 !) \times(n-4) !}=2: 1$
$\therefore \frac{\frac{n !}{(2 !) \times(n-2) !}}{\frac{n !}{(4 !) \times(n-4) !}}=\frac{2}{1}$
$\therefore \frac{n !}{(2 !) \times(n-2) !} \times \frac{(4 !) \times(n-4) !}{n !}=2$
By using above formula,
$\therefore \frac{(4 \times 3 \times 2 !) \times(n-4) !}{(2 !) \times[(n-2) \times(n-3) \times(n-4) !]}=2$
Cancelling terms $(n-4) !$ And $(2 !)$,
$\therefore \frac{(4 \times 3)}{[(n-2) \times(n-3)]}=2$
$\therefore(n-2) \times(n-3)=6$
$\therefore(n-2) \times(n-3)=(3) \times(2)$
By comparing both the sides,
$\therefore n=5$
Conclusion: Value of $\mathrm{n}$ is $5 .$
Note : Instead of taking product of two brackets in eq(1), it is easy to convert the constant term that is 6 into product of two consecutive numbers and then by observing two sides of equation we can get value of $\mathrm{n}$.